?^2»"->-"^E»
PRESENTED TO
TJBfS! CORNELL UNIVERSITT, 1870,
The Hon. William Kelly
Of Rhinebeck.
Cornell University Library
arW3908
A treatise on statics
3 1924 031 364 445
olln.anx
The original of tiiis book is in
tine Cornell University Library.
There are no known copyright restrictions in
the United States on the use of the text.
http://www.archive.org/details/cu31924031364445
A TEEATISE
STATICS,
CONTAINING THE
THEOEY OF THE EQUILIBRIUM OF FORCES;
AND
ICttrntr^ns (^um^hs
ILLUSTRATIVE OF THE GENERAL PRINCIPLES OF THE SCIENCE.
BT
S.^'EARNSHAW, M.A.
0? ST iohh's COLLXOB, caiibkisos.
FOURTH EDITION, REVISED.
DEIGHTON", BELL, AND CO.
LONDON: BELL AND DALDY.
1858.'
CamiriSae :
FBINTED BY 0. J. CLAY, M.A.
AT THE UHIVERSITY PRESS.
PREFACE TO THE THIED EDITION.
This edition of the Treatise on Statics differs so
little from the last as scarcely to require a separate
notice. A few Articles have been added, on the pres-
sure which a rigid body is made to exert on a fixed
point or axis of support by the action of forces when
there is equilibriuni. These will be found useful in
those Problems of Dynamics wherein it is reqtiired to
find the pressure which a rigid body in motion, under
the influence of any forces, exerts on a fixed point or
axis. Indeed, it is chiefly with a view to this appli-
cation of them that the articles alluded to have been
introduced into this edition of the Statics.
The collection of Problems for practice given at the
end of the Treatise has been considerably enlarged,
chiefly by the addition of Examples of an elementary
character. In the selection of them care has been
taken to choose such as illustrate Statical Principles
under every important variation of aspect, without
impeding the student's progress through them by
analytical and other difficulties foreign to the proper
object of this Treatise.
Cambkidob, Feb. i, 1845.
PREFACE TO THE SECOND EDITION.
Though the general plan and arrangement of this
edition of the Treatise on Statics are the same as in
the former, in the details there will be found, it is
hoped, some important improvements.
The fundamental proposition of the science, — the
Parallelogram of Forces, — I have proved after Du-
chayla's method, by reason of its simplicity; but I
think it necessary here to inform the reader that, as
that method is inapplicable when the' forces act upon
a single particle of matter (as a particle of a fluid
medium on the hypothesis of finite intervals), on ac-
count of its assuming the transmissibility of the forces
IV PREFACE.
to other points than that on which they act, I have,
in an Appendix, given the proof which in the first
edition was given in the text. The same objection
^and for the same reason) lies against the proof of the
parallelogram of forces from the properties of the lever.
This method, though allowable in the infancy of the
science, can never be exclusively adopted in a Treatise
which professes to take a more philosophical view of
the subject; for, were the transmissibility of force not
true in fact, the law of the composition of forces acting
on a point would «tiU be true; it is evident, therefore,
that to make the truth of the former an essential step
in the proof of the latter, is erroneous in principle.
In the former edition, forces were considered as
acting in any directions in space ; a mode of treatment
of the subject which necessarily rendered the inves-
tigations useless to such readers as had not studied
Geometry of Three Dimensions. In the present edi-
tion this defect is remedied; and a cha.pter, in which
the forces are supposed to act in a plane, is always
made to precede the more general investigations. At
the end of Chapter IV. several propositions are proved
which have hitherto been used in Elementary Books
without proof.
The fifth Chapter contains a new (and it is hoped
a satisfactory) and complete proof of the Principle of
Virtual Velocities, and its Converse. The proof given
by Lagrange in his MScanique Analytique, page 22,
et seq., though highly ingenious, I regard as a fallacy;
and, if not fallacious, deficient in generality.
In the last Chapter, I have endeavoured to set
before the reader such problems as, without involving
analytical difficulties, seemed best calculated to make
him acquainted with the mode of applying a,ll the
most important principles of the science ; and not un-
frequently I have added remarks upon important steps
with the view of pressing them more particularly upon
the reader's attention.
St John's Colleoe, CAMsniDaE,
March 12, 1842,
CONTENTS.
INTRODUCTION.
ART. PAGE
1 — 22. Definitions and Preliminary Notions 1
CHAPTER I.
23—32. Forces which act in one plane upon a particle, or upon
THE same point OF A RIGID BODV.
23, 24, Forces acting in tlie same line on a point 8
26. Parallelogram of forces. ..' 9
27] 28. Three forces acting on a point 10
29—32. Any number of forces acting in a plane on a point 11
CHAPTER II.
33 — 38. Forces which act upon a particle^ or upon the same point
of a rigid body, in any direction not in one plane is
CHAPTER III.
39 — 68. Forces which act in one plane but not upon the same
point op a rigid body.
39—57. The Theory of Couples 19
I 68 — 63. Parallel forces in a plane 29
&i — 68. Non parallel forces in a plane. 32
CHAPTER IV.
69 — 101. Forces not in one plane which act upon different
points op a rigid body.
71 — 77. Parallel forces not in a plane 36
78 — 80. Resultant of three couples 41
81 — 95. Any forces acting on a rigid body 42
96, Equilibrium of three forces acting on a rigid body. 63
98 — 101, Conditionsof equilibrium of any forces made general 64
VI CONTENTS.
CHAPTER V.
ART.
FASE
102 — 120. The Principle op Virtual Velocities 57
CHAPTER VI.
The Centre op Parallel Forces, and the Centre op Gravity.
121—126. The Centre of Parallel Forces 68
126—142. The Centre of Gravity. 70
143 — 167. Position of Centre of Gravity _ of reotilmear and regular
figures 76
168 — 173. General Properties of the Centre of Gravity 83
174 — 183. Application of Integral Calculus in finding the Centre of
Gravity of hodies 91
174, 176. Centre of gravity of a curve line —
.176. a plane area 95
177 a solid of revolution 97
178 a solid of any form 99
179 a surface of revolution 104
180 a surface of any form .106
181 , a curve of double curvature —
182, 183 hodies of variable density —
184, 186. Guldin's Properties 116
CHAPTER VII.
18G — 230. Mechanical Instruments 118
189—194. The Lever 121
195—207. The Pulley 123
208—211. The Wheel and Axle 128
212—214. The Inclined Plane 129
215. The Screw :.... 130
216. The Wedge 133
217. General Pi-operty of Machines 134
218. White's Pulley 136
219. Hunter's Screw...., ]3(3
220. Compound Wheel and Axle 137
221. The Genou.... 138
223—225. Toothed Wheels 140
.226. The Endless Screw 143
227. The Conimon Balance 144
.228. The Steelyard 146
.229. The Danish Balance I47
230. Roberval's Balance 148
CONTENTS. Vll
CHAPTER VIII.
ART. PAGB
231—236. Fbiotion 150
CPIAPTER IX.
237—244. Elastic Stbings 155
CHAPTER X.
Funicular Polygon, Catenary, Roofs, and Bridges ,,. 159
245—248. The Funicular Polygon —
249. Roofs and Bridges 161
250—267. The Catenaiy 162
CHAPTER XI.
Problems 174
Appendix 209
Miscellaneous Problems 216
By the same Author.
A TREATISE ON DYNAMICS. Tliird Edition.
STATICS.
INTRODUCTION.
DEFINITIONS AND PEELIMINARY NOTIONS.
1. In the Science of Mechanics of which Statics forms
a part, matter is considered as essentially possessing extension,
figure and impenetrability. The least conceivable portion of
matter is called a particle.
2, We conceive of matter that it can exist either in a state
of rest, or mation. If then matter, once at rest, pass into a
state of motion, the change, not being essential to the existence
or nature of ma,tter, is of necessity ascribed to some agent,
which, as to its nature, is essentially independent of the matter
influenced. Whether this agent reside in the matter influenced,
or in external objects, or in both, are questions which can only
be answered after experimental investigation. This agent is
called _^ce,' and it will be perceived from this statement, that
a force is judged of entirely by the effects which it produces :
and hence, if in the same circumstances two forces produce equal
effects, we infer that the forces are equal.
3. It is assumed, that the effect of two equal forces acting
in concert, is double the effect of one of them ; three, treble ;
and so on.
The reason of its being necessary to make this an assump-
tion is, that in our ignorance of the nature of force, we are com-
pelled to judge of it by the change which it produces in the
state of rest or motion of matter ; and it is obvious, that we can
E. s. 1
2 DEFINITIONS AND PRELIMINARY NOTIONS.
no more judge that one such change is twice as great as another,
than we can affirm ' that one candle is twice as bright, or one
substance twice as sweet, or one noise twice as loud as an-
other.
4. A force is considered as having magnitude and direction,
and a point of applicaticm. When these three are known, the
force is said to be known. From Art. 2, it will be seen that,
by the magnitude of a force, we mean the degree of motion
which it is capable of prod^icing in matter previously at rest ;
and by the direction of a force, we mean the direction in which
a particle of matter, under the influence of that force, would
begin to move ; and by the point of application of a force, we
mean that particular particle of a mass of matter on which the
force immediately exerts its influence.
5. If one particle of a rigid* mass of matter be acted upon
by a force, it cannot obey the influence of the force without
dragging with it the other matter with which it is connected ;
the motion therefore which it would receive, if free, is in some
manner distributed among the whole mass of which it is a part.
It is clear, therefore, that the subject of which we are treating,
naturally divides itself into two distinct parts, according as the
forces act on a free particle, or on a rigid body.
6. With respect to the motion of a particle of matter,
we conceive that it consists in the particle's being found to
occupy diiferent parts of space at successive instants, or epochs
of time; but with respect to the motion of a rigid body we
conceive,
(1) That as a whole it may occupy the same portion of
space at successive epochs, while some of its parts individually
occupy different parts of space in successive instants.
This is called rotatory motion.
* We define a rigid body to be an assemblage of particles of matter, connected
together in such a manner that their relative places never change.
DEFINITIONS AND PEELIMINAEY NOTIONS. 3
(2) That as a whole it may occupy different parts of space
at successive epochs, without having at the same time any
rotatory motion.
This is called a motion of translation.
(3) That both these kinds of motion may exist together in
the same body.
This is the most general kind of motion of which we can
form a notion.
7. From the preceding articles it will be perceived that we
have taken motion as the characteristic effect of force. It will
now be necessary to shew, that there exists another effect (and
that more convenient for our present purpose) which may be
taken as the measure or characteristic of force.
If any portion of matter (a stone for instance) be held in the
hand, it will be found to exert a pressure ; and if the hand be
suddenly removed, will fall. In its fall it may be caught, but
the hand will again feel a pressure. This experiment informs
us, that that which is the cause of motion, is likewise the cause
of pressure. While the stone is held at rest, its continual ten-
dency to fall is evidenced by the pressure which is exerted on
the hand ; hence, in all cases where motion is prevented, there
is pressure. But further, the latter part of the experiment
teaches us that, in all cases where motion is retarded, there is
pressure. If when the stone is at rest, the hand exert a greater
pressure upwards than is necessary to prevent it from falling,
the stone will begin to move upwards. Hence we learn that
- pressure attends the production as well as the prevention and
the destruction of motion. Thus it appears that pressure pro-
duces the same results as we have taken to be the characteristic
effects of force. We may therefore take pressure as the measure
of force, because both pressure and motion are effects of the
same cause.
8. The Earth, in some unseen manner, exerts a pressure
in a downwards direction upon all matter with which we are
4 DEFINITIONS AND PEELIMINARY NOTIONS.
acquainted. This pressure it is which occasions the descent of
falling bodies to the ground, and causes all bodies lying on the
ground to press against it. More accurate experiments prove
that every particle of matter, whether of metal, wood, earth, or
of any other substance, is subject to this influence. And it can
be shewn that the degree of this pressure exerted upon a given
body never changes. Thus, let a spring AB have one end A
firmly fixed in an immoveable block. Suspend a proposed sub-
stance P fi-om the other end B, then the spring will be bent in
the manner represented- in fig. 1, the point B taking a position
B'. If the experiment' be again tried with the same body P
after any interval of time, it will be found that the spring will
be bent exactly as at first ; thus shewing that the Earth exerts
an unvarying pressure upon every body.
If the experiment be tried with the same spring and sub-
stance P at a place in another latitude, or on a hill, or in a pit,
the bending of the spring is not found to be the same as before :
but at the same place no variation is ever observed in the
result.
9. We may easily find other substances P', P", P'"...
each of which being suspended from B will bend the spring
exactly as P does. By suspending 2, 3, 4, ... of these bodies
at a time, and marking the spaces through which the spring
is bent in each case, we may form a grxiduated scale, by means
of which we can ascertain exactly the degree of pressure which
the Earth exerts upon any proposed body whatever, as compared
with the pressure which it exerts upon P. If this be done, it
is usual to call the pressure on P the unit of pressure ; and the
pressure which is exerted upon another body, if it be TF times
the pressure on P, is said to be equal to W.
10. The pressure >F which the Earth exerts upon a body,
when measured in the manner just described, is called the weight
of the body. How great soever be the pressure which any
other force exerts upon a body, we can always find (hypo-
thetically at least) so many bodies P, P', P", P'"... that the
DEFINITIONS AND PEELIMINAEY NOTIONS. 5
Earth shall exert upon them, taken all together, exactly as
much pressure as the proposed force exerts upon the proposed
body. Hence then, with the assumption in Art. 3, we perceive
that every force may be measured, and therefore represented, by
a weight.
11. To avoid circumlocution, when a body is prevented by
an' obstacle from moving, it is usual to say that the hody exerts
a pressure upon the obstacle, and that the obstacle exerts an
equal pressure upon the body in the contrary direction. The
fact however is, that the body is completely passive ; and the
reason why it remains in a state of rest is, that it is under
the influence of two equal pressures exerted on it in opposite
directions. By the same licence, if a body, which is under the
influence of the Earth's action, be suspended by a string, it is
often said that the string exerts a force or pressure upon the
body ; the fact however in this case is, that the string by being
attached to the body, becomes a part of the body ; and the whole
remains in a state of rest, for the same reason as before. Hence
it will be seen that, in the experiment described in Art. 8, the
spring exerts a force equal to that exerted by the Earth upon P,
though in the contrary direction. And hence we say, when two
bodies are pressed together, that they act and react upon each
other with equal forces.
12. It is sufficiently evident, that two equal pressures,
acting in opposite directions upon the same point of a body,
counteract each other : but it is conceivable that if several pres-
sures be applied to a body, even though they be not two and
two in opposite directions, nor all applied to the same point of
tbe body, they may counteract each other. The science which
teaches the relations necessarily existing between the magnitudes
of forces, their directions, and their points of application, when
they exactly counteract each other, is called Statics.
13. If several forces acting upon a body counteract each
other, the body is said to be in equilibrium: and the forces are
said to balance each other.
6 DEFINITIONS AND PEELIMINAEY NOTIONS.
14. If sevieral forces acting upon a free particle do not
balance each other, the particle will hegin to move in some
direction in a certain manner. It may be prevented from so
moving by applying a proper force in the opposite direction to
that in which there is a tendency to motion. This new force
exactly counteracts the whole system of forces : but it might be
itself counteracted by a single force equal to itself and acting in
a contrary direction. A single force satisfying these conditions
would be exactly equivalent to the whole of the original system
of forces ; and it is therefore called their resultant.
15. We thus learn that several forces, if they act upon
a free particle, may be replaced by one force ; and the converse
is evidently true, viz. that one force may be replaced by a
system of several forces. When one force is replaced by a
system of several forces, they are called its com/ponents.
16. By reference to Art. 6, we see that the motions which a
rigid body may take are of two distinct kinds : and therefore the
reasoning just stated respecting a free particle does not apply to
rigid bodies. We shall hereafter shew that, corresponding to
the three cases stated in the Article referred to, a system of
forces acting on a rigid body may have
(1) A resultant for rotation only,
(2) A resultant for translation only,
(3) Two resultants, one for the rotation and one for the
translation.
17. It is evident from the explanations above given, that a
system of forces, acting on a free particle, cannot have more than
one resultant : but we have just seen that the same is not neces-
sarily true when they act on a rigid body. It is always true,
however, that the same force may have different systems of
components.
18. If a particle, or a rigid body, be in equilibrium under
the action of several forces, we may add to the system, or take
away fr-om it, any set of forces which balance each other.
DEFINITIONS AND PKELIMINARY NOTIONS. 7
This principle is called the " superposition of equilibrium,"
and we shall hereafter have frequent instances of its utility.
19. It follows at once from this, that when a body is in
equilibrium under the action of a system of forces, they may be
all increased, or all diminished in any proportion, without affect-
ing the equilibrium.
20. It scarcely needs remarking, that if a set of forces
balance each other, any one of them is equal to, and acts in an
opposite direction to, the resultant of all the rest.
21.- It is proved by experiment, that when a rigid body is
in equilibrium, any point (of the body) in the line of the direc-
tion in which a force acts, may be taken for the point of applica-
tion of the force, without affecting the equilibrium.
22. If a system of unbalancing forces acts upon the same
point of a rigid body, they will have the same resultant as if
they acted upon a free particle.
CHAPTER I.
ON FORCES WHICH ACT IN ONE PLANE UPON A PARTICLE,
OR UPON THE SAME POINT OF A RIGID BODY.
23. To find the resultant of several forces acting, in the same
line, wpon the same point of a rigid hody.
If all the forces act in the same direction along the line,
they will produce the same effect as a single force equal to their
sum.
If some act in one direction and some in the opposite direc-
tion, then by the first case the resultant of each set wiU be equal
to the sum of the forces of which it is composed : and these two
resultants, acting in opposite directions, will be equivalent to a
single resultant equal to their difference. Hence then whether
the original forces act in the same or in opposite directions, their
resultant is equal to their algebraic sum.
In forming this sum, we are to account those forces positive
which act in one direction, and those negative which act in the
opposite direction ; and the algebraic sign of the sum so formed
will shew in what direction the resultant acts.
24. Cor. If a number of forces act in the same line upon
the same point of a rigid body, they will be in equilibrium when
their algebraic sum is equal to zero, for in that case their re-
sultant vanishes, and they produce no effect. Hence the condi-
tion of equilibrium of any number of forces acting in the same
line and upon the same point of a rigid body is that their alge-
braic sum shall be equal to zero,
25. Dep. Lines are said to represent forces in magnitude
and direction, when they are drawn parallel to the directions in
which the forces act, and have their lengths proportional to the
magnitudes of the forces.
FORCES ACTING IN A PLANE ON A POINT. 9
26. If from a point two lines he drawn representing two
forces which act upon a point; and if upon these lines a parallel-
ogram he constituted, the diagonal drawn from the same point
will represent the resultant of the two forces. This property is
usually cited as "the parallelogram of forces."
We shall first prove that the diagonal represents the direction
of the resultant force. This part of the proposition is evidently
true when the two given forces are equal ; let us assume that p,
q and r are three forces, such that this is true for p and q ; and
also for jj and r. At the point A, fig. 2, apply p in the direction
AB: and g-, r, both in the direction AD. Take AB, AC, CD
to represent the respective magnitudes of these forces. Complete
the parallelograms, and draw the diagonals as in the fig. The
resultant of p and q acts in the direction AE, hy hypothesis ;
and we may by Art. 21, suppose it to act at E; and we may
there resolve it into its original components p and q ; the latter
acting in the line EF, and the former in the line GE produced ;
this we may suppose hy Art. 21 to act at 0, and the former at
F; also the force r may he supposed to act at G. We have now
two forces p, r &t C represented hy the lines GE, GD ; their
resultant by hypothesis acts in the line GF, and therefore we
may suppose at F. The three forces p, q, r, which originally
acted at A, are by this process reduced to forces acting at F. F
is therefore in the line in which their resultant acts when they
are applied at A, (Art. 21). lfow-4Z), AB, represent twoibrces
q-\-r and p ; and we have just shewn that their resultant acts in
the direction of the diagonal AF. If then our proposition be
true for the two forces p, q ; and also for the two forces p,r; it
is also true for the forces^ and q-\-r. Now it is true when^, q,
f are all equal ; and hence it is true for ^ and 2p : and because it
is true for p, p; and also for ^, 2p; therefore it is true for^,
3p;... and by following the same mode of reasoning it is true for
p and mp, m being any whole number.
Again, because it is true for mp and p, and also for mp
and p ; therefore it is true for mp and 2p ; and as before for
mp and np, n being an integer. Hence our proposition respecting
the direction of the resultant is true for any two commensurahle
forces mp, np.
E.S. 2
10 FOKCES ACTING I.N A PLANE ON A POINT.
If the proposed forces P, Q be incommensurable, by taking p
extremely small and the integers m, n correspondingly large, we
can make mp differ from P, and np from Q by less than any
quantities which can be assigned ; and we may then use mp and
np, instead of P and Q, without any sensible error ; and there-
fore the proposition is true of P and Q.
We shall now prove that the diagonal represents the magni-
tude of the resultant force.
Let A C, AB (fig. 3) represent the magnitudes and directions
of two forces acting on a point. Complete the parallelogram : its
diagonal ^^ has been proved to represent the direction of the
resultant ; it also represents its magnitude. For in EA produced
take ^i^to represent its magnitude; then AB, AG, AF repre-
sent three forces which balance each other : wherefore completing
the parallelogram AFQB, its diagonal A Q represents the direc-
tion of the resultant of AF, AB, and is consequently in the same
■ line with A G (Art, 20). Hence A GBE is a parallelogram, and
therefore ^JS^= OB = AF; that is, ^^ represents the magnitude
of the resultant oi AB, AG.
27. If three forces acting on a point are represented hy
the sides of a triangle taken in order, they will halance each
other. And conversely ; If three lines, forming a triangle, he
parallel to the directions of three forces which, acting on a point, ~
halance each other, the sides of the triangle taken in order will
represent the forces.
For let AB, BE, EA (fig. 3) represent the forces P, Q, B
which act on a point. Complete the parallelogram BG; then
because ^ C is drawn parallel and equal to BE, it also represents
the force Q. Hence the resultant of P and Q is represented by
AE; which being equal, and in a contrary direction, to EA
which represents R, there is equilibrium.
Conversely, let the three forces P, Q, B, acting on a point
balance each other : and suppose ABE (fig. 4) to be the triangle
whose sides are respectively parallel to the directions in which
P, Q, B act. Two, at least, {AB, BE suppose) represent the
directions of the corresponding forces ; and we are at liberty to
FORCES ACTING IN A PLANE ON A POINT. 11
suppose that one of these {AB suppose), represents also the
magnitude of its force P: if BH do not represent the magni-
tude of the other force Q, take BJE' to represent it, and join
AE'. Then (by the former case) B,'Q and a force represented
by U'A will balance each other. But F, Q, R balance each
other; and therefore R is represented by E'A both in magni-
tude and direction; which is impossible (because EA repre-
sents R in direction by hypothesis) unless E' coincide with E.
Therefore E' does coincide with E, and therefore the forces are
represented by AB, BE, EA, which are the sides of the triangle
taken in order.
28. If three forces, acting upon a point, balance each other,
their directions lie in a plane ; and their magnitudes are respec- ■
tively proportional to the sine of the angle between the directions in
which the other two act.
Let the forces be P, Q, R (fig. 5) acting in the directions
OA, OB, 00. In OA, OB, take points A, B, such that OA,
OB represent the magnitudes as well as the directions of P,
Q. Complete the parallelogram A OBI), and join OB. Then
Q being represented by OB may also be represented by AB;
also as the three forces represented by OA, AB, BO, acting on
a point will balance each other (Art. 27), therefore P, Q and
a force represented hy BO balance each other; but P, Q, R
balance each other ; therefore R is represented hj BO: and con-
sequently BOG is a straight line, and OA, OB, 0(7 lie in the
plane of the triangle OAB. Also
P : Q : R :: OA : AB : BO
sin OB A : sin BOA : sin OAD
sin BOB : sin AOG : sin PAB
sin QOR : sin POR : sin POQ,
''' ^^ioR'mSoR^^KTOQ- Therefore, &c.
29. Two forces act upon the same point in directions at right
anghs to each other, to find their resultant (fig. 5*).
12 FORCES ACTING IN A PLANE ON A POINT.
Let tbe forces be X, Y acting upon the point in the direc-
tions Ox, Oy at right angles to each other. Take OM, ON to
represent the forces ; complete the rectangle OMPN, and draw
the diagonal OP. This line by Art. 26 represents the resultant
of X and F. Let B be the resultant, and 6 the angle POM
which its direction makes with "the direction of the force X
Now OP'=OM' + MP';
.■.P?= X'+Y',
which determines the value of B : and then the equation
. MP Y
^^^ = -OM^X
determines the value of 9.
30. COE. If a force B be given and it be required to
resolve it into two components acting in directions at right angles
to each other, we must employ the equations
X = 5 cos e, and r= ^ sin 6,
which are derived from the equations
0M= OP cos 6, and MP= OP. sin 9.
31. Any nwmher of forces act wpom, a point in given directions
in a plane; to find their resultant.
Let F^, F^, F^...F„ be the forces, and (see fig. of Art. 29)
the point upon which they act. In the plane in which are the
lines in which the forces act, draw any two lines Ox, Oy at right
angles to each other; and denote by a,, a^, ag...a„the angles
which the directions oiF^, F^, j?'3...i^„ respectively make with Ox.
Then the components of these forces are respectively (by
Art. 30),
F^ cos «! , F^ cos ttj, jPj cos Kg F„ cos a„
in the direction Ox ; and
F^ sin ftj, F^ sin a^, F^ sin ag F^ sin a^
in the direction Oy.
Let us replace (see Art. 15) the original forces
FORCES ACTING IN A PLANE ON A POINT. 13
by tliese two sets of componentg. These components are re-
spectively equivalent to two forces acting in the lines Ox, Oy
(Art. 23), and being equal to
F^ cos tti + F^ cos a^ + -PI, cos Kg + ... + -F„ cos a„,
and F^ sin a, + F^ sin a^ + F^ sin a, + . . . + i^„ sin a„.
Let B be the result of the original forces
F F F . . F
and suppose that 6 is the angle which the line in which it acts
makes with Ox. Then since B is equivalent to the original
forces, it is also equivalent to the two components of them which
have just been found : hence (by Art. 30)
B cos6 = F^ cos Kj + i^2 cos a^ + ... + F„ cos a„,
and B sm6 = F^ sin a^ + F^ sin a.^+ ... +F„ sin a„.
From which two equations both B and 9 may be found.
Eemaek. The sum of a number of quantities of the same
form is oftMi, for brevity, represented by prefixing the symbol
S to a term representing the general form. Upon this principle
the above equations may be written thus :
B cosd — %{F cos a), ox%.F cos a,
and i? sin = X (i^ sin a), or S . i'' sin a ;
:. B?={X.FcoaoLY+{t.F sin a)»,
S.-Fsina
and tan 6 =
% .Fcosol'
32. To find the conditions of equilihrium of forces acting
upon a point in any directions in one plane.
Let F^, F^... Fnhe, the forces ; Oj, a, ... a„ the angles which
their directions make with a line Ox ; then, proceeding as in the
last article, We have the equations there found for the determi-
nation of their resultant. But because they balance each other
by hypothesis, they have no resultant, and therefore .H = 0, or
= (S . i^ cos a)'' ->r{t.F sin a)^
14 FORCES ACTING IN A PLANE ON A POINT.
But as the right-hand member consists of two terms which, being
squares, are essentially positive, their sum cannot be equal to
unless each be separately equal to ;
.•- Q = 't.Fcosa, and = S.i^sina.
If the investigation in Art. 31 be examined, it will be seen
that the line Ox was taken in any direction in the plane of the
forces ; and hence we may in the most general terms state the
signification of the two equations just found to be as follows,
If any number of forces act in one plane upon a point, that
there may be equilibrium.
The sums of the components of the forces parallel to any two
lines, at right angles to each other, in the plane of the forces must
he separately equal to zero.
The converse is evidently true also.
No other condition is necessary for equilibrium, for if
2.i'^cosa=0, and S.i^sina=0, it follows inevitably that
J? = 0, and therefore there is equilibrium.
CHAPTEE II.
ON FORCES WHICH ACT UPON A PARTICLE, OR UPON THE
SAME POINT OF A RIGID BODY, IN ANY DIRECTIONS NOT
JN ONE PLANE.
33. If three forces acting upon the same point he respectively
represented hy the three edges of a parallelopiped which meet, the
diagonal of the parallelepiped drawn from, that point to the
opposite corner will represent their resultant.
For let OA, OB, OG (fig. 6) be the edges wliicli represent
the three forces, and OE the diagonal of the parallelopiped :
draw OD, GE.
Then because OA, OB represent two forces, OD represents
a force which is equivalent to them both (Art. 26): hence the
three forces represented by OA, OB, G are equivalent to the
two represented by OD, OG, which again are equivalent to the
single force represented by OE, for GD is a parallelogram.
34. Three forces act upon a point in directions which are
at right angles to each other; to find their resultant.
Let X, Y, Z be the forces, acting upon the point (fig. 7)
in the lines Ox, Oy, Oz which make right angles yOz, xOy,
z Ox with each other. From set off OL, OM, ON to represent
the forces X, Y, Z respectively. Complete the parallelograms
OMQL, OQPN, and join OP; this line, by the last Art., repre-
sents the resultant required.
Let B denote the resultant, and a, /3, 7 the angles POx,
POy, POz which its direction makes with the directions of
the given forces.
16 FORCES NOT ACTING IN A PLANE ON A POINT.
Then because 0N= OP. cos 7 ;
.•. Z= ^ cos 7 1
Similarly F = iZ cos ^ i . . . (A) ;
and X=B cos a)
.: X'+Y^ + Z' = B' (cos' a + cos' /3 + cos' 7)
/OX' ow oir\
~ [OF"^ Or^ OF')
_ ™ OL' + LQ'+QP' _ j^ OQ'+QJP'
-^ Oi« ~ OF'
OF'
'OF'
This equation gives the value of B, and then the three
equations marked (A) give the angles a, /8, 7, which fix the
line in which B acts.
Remark. The reader will observe from the above that
cos' a. + cos' /8 + cos' 7 = 1,
a formula which will be of frequent use in the following pages.
35. CoE. If a force B be given, and it be required to
resolve it into three components, whose directions are at right
angles to each other, we must employ the equations marked (A).
36. Any number of forces act in given directions upon a
point; to find their resultant.
Let (fig. 7) be the point upon which the given forces
Fj^, F.^, F^ ... Fn act; from draw three lines Ox, Oy, Oz,
arbitrarily taken, at right angles to each other ; and denote by
«A7u aA^a) «A7s ••• «A7„, the angles which the directions
of the forces make with these three fixed lines.
The respective components of the given forces are
-F, cos a,, F^ cos a^, ... F^ cos a„,
in the direction Ox;
POECES NOT IN A PLANE ACTING ON A POINT. 17
F, cos^., F^ cos/3,, ... F„ cos ,S„,
in the direction Oy; and
F^ cos 7, , F^ cos 7j, . . . F^ cos 7„,
in the direction Oz.
Replace now the original forces hy these three sets of com-
ponents (Art. 15) ; each set is reduced to one force by Art. 23 ;
and we then have three component forces
2 . i^ cos a, % . F cos /S, % . F cos 7,
acting respectively in the lines Ox, Oy, Oz.
Let R be the resultant required, and a', yS', 7' the angles
which the line in which it acts makes with Ox, Oy, Oz. Then
since R is equivalent to the original forces,^ it is also equivalent
to the three components of them which have just been found ;
hence
R cos a' —%. F cos 0L'\
^ cos jS' = 2 . i^ cos/si ... (A) ;
R cos 7' = 2 . i^cos 7'
and therefore since 1 =cosV+ cos'/S' + cosV) (-A^t- 34, Rem.)
we find
B' ={t.F cos a)" + (2 . ^ cos ^Y +{%.F cos 7)'.
This equation gives the value of R ; and then the equations
(A) will give a', /3', 7', which fix the direction in which R acts.
37. To find the equations of the line in which the resultant
a/its.
Suppose OP (fig. 7) the line in which the resultant acts ;
and let x, y, z be the co-ordinates of any point P in it.
Then if OP be taken to represent R, the co-ordinates will
represent the components, and therefore by Art. 25,
X _ y _ z
2 . i^cosa~2 . i^cos/3~2 . -FC0S7'
which are the equations required.
E. s. 3
18 FORCES NOT IN A PLANE ACTING ON A POINT.
If the point at whicli the forces act be not the origin of
co-ordinates, let its co-ordinates be a,h,c; then since the line
whose equations are required passes through this point,
x—a _ y—h e—c
STFcosa ~ sTTcos^S ~ S . -F cos 7
are the equations of it in this case.
38. To find the conditions that a system of forces acting
upon a point in any directions, may balance each other.
It is evident that there cannot be equilibrium among the
forces F^, F^, F^ F„, unless their resultant be eyanescent,
and therefore we must have
= (2 . ^ cos a)'' + (S . i^ cos /3)= + (S . i^ cos 7)',
which for a reason similar to that assigned in Art. 32 resolves
itself into the three independent conditions
= S.^cosa, = S.i^cos/3, = 2.^ cos 7.
Or, in words, (remembering that the positions of Ox, Oy, Oz
were arbitrarily chosen).
The sum of the components of the given forces parallel re-
spectively to any three lines at right angles to each other must
separately he equal to zero.
CHAPTEK III.
ON FORCES WHICH ACT IN ONE PLANE BUT NOT UPON THE
SAME POINT OF A RIGID BODY.
THE THEORY OF COUPLES.
39. Remark. It has been stated in Art. 21, that the effect
of a force is not altered by supposing it to be transferred from
one point of the body in the line of the direction of its action
to another: from this it follows that if the directions of the
forces which act at different points of a rigid body, all pass
through a point, we may fictitiously transfer them to that point,
and then by the preceding Chapters find their resultant, which
in its turn we may transfer to any convenient point of the rigid
body which happens to lie in the line of its direction. It is
obvious, that when any two forces in the same plane act upon
a rigid body at different points, their directions unless parallel
being produced will meet, and therefore after the statement just
made it will not be necessary to include the consideration of
two non-parallel forces in the present Chapter, we shall there-
fore begin with the following.
40. Two forces act in 'parallel directions upon different points
of a rigid hody, to find their resultant.
Case 1. Let F, F' be the two forces, and let us, first,
suppose them to act in the same direction.
Let A, B (fig. 8) be any two points of the rigid body in the
lines of direction of the respective forces: join A, B; at these
points in opposite directions along the line AB apply any two
equal forces /, /'. These being in equilibrium produce no
effect.
20 THE THEORY OP COUPLES.
Now i^ and/ (by Art. 26) and F" and/' will have resultants
{rn,, n suppose) acting in certain directions Am, Bn within the
angles FAf, and F'Bf: these lines being produced will meet in
some point P to which let m, n be transferred : and let them
there be resolved into their original components; viz. m into
/ and F, acting at P in the directions Pf and PR {PR being
drawn parallel to AF) ; and n into/' and F' acting at P in the
directions Pf and PR, which is also parallel to BF'. The
forces /and/' at P being in equilibrium may be removed, and
there remain the original forces F, F' both acting at P along
the line PR parallel to their direction at A and B. Hence the
resultant of F and i^ is a force, equal to their 'sum F-\- F',
acting at any point in the line PR ; the position of which we
find as follows.
Let PR cut AB in Q. Then because m is the resultant
of P and/ a force equal to m applied at ul in the direction AP
would keep the two forces P,/in equilibrium; and the three
being paa-allel to the sides of the triangle APQ taken in order,
aie proportional to those sides (Art. 27) ;
.-. P : / :: PQ : AQ.
Similarly /' : P':: BQ : PQ
.: F:F::BQ: AQ; •.•/=/'.
Consequently Q divides AB into two parts which are inversely
proportional to the forces adjacent to which they lie.
41. Case 2. Let us now suppose the two forces P, P' to
act in contrary directions, as in fig. 9, and that they are unequal,
P being the greater.
Introduce the equal and opposite forces /and/' as before;
and let m be the resultant of P and/; and n that of P' and/'.
Then since the angle FAf is equal to the angle F'Bf, it will be
found, by constructing the parallelograms of force upon FA, Af,
and F'B, Bf, that since P is greater than P', the direction
of the resultant m lies nearer to AF, than the direction of the
resultant n to BF: that is, the angle fAm is. greater than the
angle fBn or ABP. Consequently the lines nB, Am being
produced will meet on the side towards the greater force P
THE THEORY OF COUPLES. 21
as is represented in the figure. From this point proceeding as
in the former case we find that the forces F, F' preserving their
proper directions, may be removed to the point P. Hence their
resultant R is equal to i^— F', the algebraic sum of the forces,
and acts in the direction of the greater force. The word sum is
used in the statement of this result, because jP being assumed
positive, F acting in the contrary direction must be accounted a
negative force. (See Art. 23.)
The position of the point Q is found as before from the
proportion
F : F' :: BQ : AQ,
and it is to be noticed particularly thfit Q lies in BA produced ;
and is situated nearer to A (the point of application of the
gi-eater force) than to B.
42. Case 3. Let us lastly sujypose the two forces F, F'
acting in contrary directions, to he equal (fig. 9).
In this case the angles fAm, f'Bn are equal ; and conse-
quently the angles fAm, ABP are cqnal, and the lines Am, nB
are parallel, and have no point of concourse. It would appear
then, that the former mode of finding the resultant of F and F'
fails entirely in this case. The present case may, however, be
considered the ultimate state of Case 2, at which we arrive
by supposing the magnitude of F' to approach continually
nearer to that of F, until at length their difierence becomes less
than any assignable quantity. Let us then reconsider Case 2.
We have found
B = F-F',
and F : F' :: BQ : AQ;
■•'BQ-^ • a)
or R . BQ = F. AB. (2).
Hence, we see from (1) that as F' increases, the point Q
moves continually farther from B, and BQ becomes infinite in
the ultimate state; and at the same time from (2) we see that
the resultant R diminishes in such a manner that the product
22 THE THEORY OF COUPLES.
B.BQ never changes; and B becomes zero in the limit. Hence,
in the ultimate state, that is, when F' differs from J?' by less than
any assignable quantity, we have a resultant zero acting at
a point infinitely distant from ^ or -B; yet even then the
product B.BQ remains finite, which apart from any other
consideration would induce us to conjecture, that some finite
effect is due to the action of jFand F' in this case, although not
such an one as can be represented by a single force.
Upon these grounds we conclude, that a system of two equal
forces acting in contrary directions on different points of a rigid
hody is not reducible to a single resultant.
43. Deps. a system of two equal forces acting upon a
rigid body in opposite directions but not in the same straight
line, is denominated A couple. A plane which passes through
the two lines in which the forces of a couple act, is called the
plane of the couple.
When the line AB (fig. 9) is drawn at right angles to the
directions of the forces of the couple, it is called the arm of the
couple; and the product F . AB (see (2) of Art. 42) is then
called the moment of the couple.
44. E.EMAEK. It is obvious from an examination of fig. 9,
that one effect of a pair of forces, acting in contrary directions
at different points of a rigid body, whether they be equal or
unequal, is the communication of a rotatory motion (see Art. 6)
to the body on which they act ; what other effect they would
produce is not so obvious, nor indeed does it belong to us, in
treating of the present subject, to consider what is the effect of
unbalanced forces in any case. For the satisfaction of the
reader, however, and for convenience in what follows, it may be
stated, that it is proved in Dynamics, that the sole effect of a
couple is to communicate to the body on which it acts an
angular motion about an axis passing through a certain point in
the body, called the centre of gravity.
45. Def. If the forces of the couple act .so as to tend to
turn the body round in the direction of the motion of the hands
of a watch, it is called a right-handed couple, and more fre-
quently a positive couple ; but if, as in fig. 9, the forces act so as
THE THEORY OP COUPLES. 23
to turn the body in the contrary direction, the couple is styled
left-handed, or negative.
These terms, to prevent confusion, will be used in this book
as here defined ; but the reader will observe that, in Statics as
in Algebra, the terms positive and negative are only relative,
and may be applied, at discretion, to any two forces acting in
contrary directions, or to any two couples which tend to com-
municate opposite angular motions to the body on which they
act.
46. The reflecting reader will have remarked that a couple,
though positive when viewed by a spectator looking at it from
one position, appears negative to a spectator looking at it from
a position on the other side of its plane. A couple is therefore
positive or negative, according to the situation of the spectator,
with respect to its plane. It will prevent confusion, if we call
that face of the couple's plane the positive face, upon which the
spectator looks when the couple appears to him to be positive:
the other face of the plane must then be considered negative.
47. Deps. a straight line, in length proportional to the
moment of a couple, being drawn perpendicular to the plane of
the couple, is called "the axis of the couple"
And it is said to be the jpositive, or the negative axis, ac-
cording as the perpendicular stands on the positive, or on the
negative face of the couple's plane.
If the axis of a couple is mentioned without its being stated
whether it is positive or negative, we are to understand that the
positive axis is alluded to.
The angle between the planes of two couples is measured by
the angle between their positive axes.
48. The effect of a couple acting on a rigid body is not
altered hy twrning its a/rm through any angle in the plane of the
couple.
Let F and F' be the equal forces of a couple acting at two
points in the lines FA, F'B (fig. 10), and having the arm AB.
From A, any point in the line in which F acts, draw in an
arbitrary direction in the plane of the couple AB' equal to AB;
24 THE THEORY OP COUPLES.
and at A, B' in the plane of the couple F, F', and in directions
at right angles to AB' apply two pairs of opposite fotces /, g;
f, g': each force being equal to JP'or F'. These being in equi-
librium, produce no effect.
Bg' and F'B^W intersect each other in some point G; join
AG. Then because AB==AB', and AG is common to the
triangles ABO, AB'G, and the aB= aB'; therefore ^C bisects
the angles^ GS', BAB': hence the resultant of the two equal
forces F' g" which we may suppose to act at G lies in AC
produced ; and that of the equal forces F, g lies in GA pro-
duced. But, because AF is parallel to GF' and Ag parallel to
Of, therefore the iFAg= iF'Gg'; and consequently the re-
sultant of the forces F', g" is equal to thait of the forces F, g :
we have just proved also that they act in opposite directions ;
therefore, the four forces i?; ^, i^', / balance each other, and
may be removed.
There is then left only the couple/,/', which is the same as
if the arm of the original couple had been turned through the
arbitrary angle BAB'.
49. The effect of a cowple acting on a rigid body is not
altered hy removing it to any other part of the rigid hody\ pro-
vided its new plane he parallel to its original ^Icme (fig. 11).
Let F, F' be a couple acting upon a rigid body in the plane
HH; let AB be its arm. Let KK be any other plane, in the
rigid body, parallel to HH; and in this plane draw the line ah,
parallel and equal to AB. At a and h apply pairs of opposite
iovcea f g; f, g': each force being equal and parallel to the
forces F, F'. These pairs of forces balance each other, and
therefore produce no effect on the rigid body. Draw Ah, Ba ;
they, being in the plane which contains AB and ah, necessarily
intersect in some point 0. In fact, if A, a, and B, h were
joined, AahB would be a parallelogram, and therefore Ah, Ba,
being its diagonals, bisect each other in 0. Draw PGQ parallel
to AF. Then because F=: g', and & C = ^ 0, ' '
.-. F : g' :: hC : AC.
Hence Fani g! (by Art. 40) have a resultant F+g', which
acts at in the line GP. Similarly it may be shewn, that F'
THE THKOKY OF COUPLES. 25
and g have a resultant, F' -\- g acting at C in the line CQ.
Now F+g' = F' +g and CP is opposite to CQ, therefore the
four forces F, g", F', g balance each other, and may he removed.
There remains then only the couple f, f, which is the same as
if the original couple had been removed into the new plane
KK, retaining its arm 06 parallel to -^5; but we may now (by
Art. 48) turn the arm ab through any angle without altering the
effect of the couple. And hence the effect of a couple, &c.
50. The effect of a couple acting on a rigid hody is not
altered hy removing it to any other part of the rigid hody in its
own plane.
The demonstration in the last Article will serve for this,
using fig. 12 instead of fig. 11.
51. A Gowple acting on a rigid hody may he changed for any
other cowple acting upon the same rigid hody, provided the moments
of the two couples he equal, their planes parallel, and they he both
of the same hind, i. e. hoth positive or hoth negative.
Let HE (fig. 11) be the plane of the couple F, F' ; and in
any other plane KK oi the rigid body draw, parallel to AB,
a line cd> of any proposed length : at a, h apply pairs of equal
and opposite forces/, g; f, g'; of such magnitude that
F.AB=f.ab,
these balance each other, and therefore produce no effect.
Now AB and ah being parallel, the lines Ah, Ba lie in one
plane, and intersect in some point C: and because AB i^ parallel
to ah, the ^ GAB= ^ Cha, and the ^ CBA = ^ Cab ; consequently
the two triangles AGB, hca are similar. Now because /=^'
.-. F : g' :: ab : AB
:: Ch: CA;
therefore, by Art. 40, the resultant F+ g' of the two forces F, gf
acts at C in the direction GP. In a similar way it may be
shewn that F' +g, the resultant of F', g, acts at C in the direc-
tion CQ. Now F= F' and g'=g, and therefore F^g'=-F'^g;
consequently the four forces F, cf, F', g are in equilibrium, and
may be removed; which being done, the original couple is
E. s. -4
26 THE THEORY OF COUPLES.
replaced Iby the equivalent couple f, f whose arm is ah. This
couple/,/' may now he turned through any angle in the plane
KK, and thus the proposition is established.
52. Any number of couples act upon a rigid body in the same
plane, or in parallel planes ; to find their resultant.
Change all the couples into others equivalent to them, and
therefore of the same moment, but all having their arms of the
length h. Then if F^,F^, F^...F^ be the forces; and a,, a^, a^...a^
the arms of the original couples; and P^, P^, P^,...Pn the forces
of the corresponding equivalent couples, we shall have
PJ>=F^a„ P,b = F^a^,...PJ = F„a,.
Now since the new. couples act in parallel planes and have
equal arms, they may be removed into the same plane, and then
turned round and transposed so as to make all their equal arms
exactly coincide; in which position the system of couples is
reduced to one couple, the arm of which is b, and the forces of
which are equal to
F, + P, + P,+ ...+P^.
Hence the moment of the resultant couple
= (P, + P, + P3+...+P„).J
= F,a, + F^a, + F^a,+ ... + F„a„
= the sum of the moments of the original couples.
Whence, the moment of the resultant couple is equal to the sum
of the moments of the original couples.
The reader will be careful to remark, that if any of the
couples are of a negative character, their moments are to be
accounted negative in taking this sum.
53. COE. If all the n couples be equal, the moment of
their resultant couple is n times the moment of one of them;
THE THEORY OP COUPLES. 27
and as the effect of n equal couples must he n times the effect of
one of them, it follows that the moment of a couple is a proper
measure of its effect in producing or destroying equilibrium.
Whenever, therefore, we have occasion to speak of the magnitude
of a couple, we shall do so by stating its moment ; thus, the
couple G will signify the couple whose moment is G. It will,
lead to no inconvenience that the magnitude of the forces which
compose the couple is not stated, seeing that the effects of all
couples of equal moments acting in the same plane, whatever be
the magnitudes and directions of their forces, are the same. It
will also be observed, that it is not necessary to state the precise
plane in which a couple is situated ; it will be sufficient to know
its moment, and the position of some line to which its axis is
parallel.
54. It will be observed, that all equivalent couples have
their axes equal and parallel.
55. If from a point two straight lines be drawn parallel and
equal to the axes of two couples, and upon them a parallelogram
be described, the diagonal drawn from the same point will be
parallel and equal to the axis of the resultant couple. (This pro-
position is usually cited as the parallelogram of couples.) (fig. 13).
As the planes {SO A, HOB, suppose) of the couples are not
parallel, let them intersect in the line HO, Change the couples
into two equivalent couples having their forces FF', ff' all
equal; place these new couples so that one extremity of their
arms OA, OB shall be at 0, and the forces F, f which act
there, shall act in the line OH, as in the figure. Complete the
parallelogram OADB, and draw the diagonals OD, AB, bisecting
each other in C. Then because F' and /' are equal and act in
the same direction, they are equivalent to a resultant F' +f'
acting at" G (Art. 40). But such a force at G would likewise be
the resultant of the same forces F', f acting at B, 0. We may
therefore transpose F' to B, and /' to 0, which being dtone,
/' and f s.t balancing may be removed ; and there will only
remain Fat and F' at B, forming a couple F, F' whose arm
is OB, which is therefore the resultant of the two original
couples.
28 THE THEORY OP COUPLES.
Now the forces of the two component couples and of their
resultant being equal, their axes which are proportional to their
moments, are in this case proportional to their arms OA, OB,
OD; we may therefore consider OA, OB, OB as being equal to
the axes. If therefore from in~ the plane OBA, we draw
■ three lines respectively perpendicular and equal to OA, OB, OB,
they will be the axes of the three couples, and will then have
the same position as the lines OA, OB, OB would take if the
parallelogram OABB were turned through a right angle about
the fixed line OE. This figure OABB so turned is the paral-
lelogram stated in the enunciation of the proposition to possess
the property which we have just proved belongs to it.
56. Two couples act upon a rigid hody in planes which are
at right angles to each other ; to find their resultant.
From any point 0, draw OA, OB
equal and parallel to the axes of the two
couples. Complete the rectangle OACB,
and draw its. diagonal. OG. By the last
article 00 is equal and parallel to the
axis of the resultant couple. .
Let L, M, G be the moments of the two component couples
and of their resultant. 6 = GO A the angle at which the axis of
G is inclined to that of L.
Then because OA=OG cos 0, and OB=OG sind;
.'. L= G cos 6, and M= G sin 6,
from which we find
G' = L'+M',
and tan ^ = -f- ;
which equations determine both the magnitude of the resultant
couple, and the position of its axis.
57. If it should be required to resolve a given couple whose
moment is G into two components acting in planes at right
angles to each other, we must use the equations
L=Gcoa9, M=Gs[n0.
PAEALLEL FORCES IN A PLANE. 29
58. Any number of forces act on a rigid body in parallel
directions in one plane at different points of the body; to find their
Let F^, F^... F^hQ the forces ; from any point (fig. 14)
of the rigid hody in the plane of the forces draw a line
cutting their directions perpendicularly in the points A, B ...H;
and put
OA = x^, OB=x^,...OH=x,.
At apply two opposite forces each equal and parallel to F^ ;
they do not afiect the system. In the same way apply at
a pair of forces for each of the remaining forces F^, F^...F„.
By this means we have n forces acting at in the direction OR,
respectively equal to F^, F^, ... F^; these are equivalent to a
single force B, acting at in the direction OB, and
B = F^ + F^+... + F^^t.F.
We have, besides, n couples whose arms are a;,, x^...x„, which
are (by' Art. 52) equivalent to a single couple, whose moment
= F^x^ + F^x^+...+F^x^
= %.Fx.
Consequently the given system of forces is equivalent to a single
force % . F acting at in the direction OB ; and a negative
couple whose moment is S . Fx.
59. The result just obtained is perfectly general, but admits
of simplification except in the particular case when 'St .F=0.
(1) In the particular case when "t.F^O, there is no re-
sultant force acting at 0, and therefore the only resultant is the
couple whose moment is "Z . Fx.
(2) When "Z .Fia not =0, change the couple whose moment
= 't.Fx into an equivalent couple which has its forces B', B"
equal to ^ ox't.F, and place it so that its arm OK (fig. 14)
shall coincide with the line OH;
.-. f^.F).0K^t.Fx (Art. 50.)
30 PARALLEL FORCES IN A PLANE.
By the arrangement the force i? at is balanced -by B" one
of the forces of our new couple ; and these being removed, there
remains only the force B' = % .F&t the point Z" determined by
the equation
t.Fx
0K--
t.F
Consequently, except when S . i^= 0, the resultant is a single
force equal Xo%iF acting at the point just found.
60. COE. If the line OH instead of cutting the directions
of the forces at right angles, should cut them in an i a, we
should have found that
(1) "When % . F= 0, the resultant is a couple whose moment
is {% . Fx) sin a : and
(2) When S . i^ is not = 0, the resultant is a single force
S . F acting at the point determined by the same equation as
before, viz.
t.Fx
0K=
t.F
61. Any numh&r of parallel fc/rcm act upon a rigid lody in
one plane at different points of the lody ; to find the conditions
that they may balance each other.
Let the system of forces be that of Art. 58 ; then we have
to consider the two cases pointed out in the last Article. In the
second case the resultant is the force S . ^acting at K; and there
cannot be equilibrium unless this force vanish, or % . F= 0, But
if this be the case, the second case coincides with the first ; and
the resultant is a couple whose moment = % . Fx : there cannot
be equilibrium therefore unless this couple also vanish. Conse-
quently the conditions of equilibrium are
t.F=0, and.t.Fx = 0;
these are both necessary and sufficient for equilibrium. They
are necessary, for if the former only be satisfied, there will exist
the couple of Case 1 ; and if the latter only be satisfied, there
will exist the resultant force acting at 0. And they are
NON-PARALLEL FORCES IN A PLANE. 31
sufficient, for they secure that there shall exist neither the re-
sultant of Case 1, nor that of Case 2.
62. Def. The products F^x^, F^x^, ... F^x^ are called the
moments of the forces F^, F^, ... F^ about the point ; they are
also called the moments of the same forces about an axis passing
through at right angles to the plane of the forces.
Hence remembering that the point was arbitrarily chosen
in the plane of the forces, the two conditions of. equilibrium of
parallel forces acting on a rigid body in one plane may be thus
enunciated in words : —
The algebraic sum of the forces, and the sum of the moments
of the forces about any point in the plane of the forces or about
any axis perpendicular to the plane of the forces, must be each
equal to zero.
63. Suppose that there is in the plane of the forces a fixed
point, or in the body a fixed axis not parallel to the plane of the
forces: to find the conditions of equilibrium.
If there be a fixed point In the plane of the forces, let that
.point be taken for 0; or if there be a fixed axis it will cut the
plane of the forces in a point, which take for ; then the in-
vestigations of Art. 58 apply here. The force % . F which acts
at 0, can produce no effect since it acts on an immoTeable point ;
it is not necessary therefore that % . F should be = 0. But the
couple whose moment = S . Fx, if it exist, will turn the body
about 0; and therefore that there may be equilibrium, it is
necessary and suflScient that
t.Fx = 0;
hence there is only one condition of equilibrium in this case,
which we may thus express in words : —
The sum of the moments of the forces about the fixed point,
or about that point where the fixed axis cuts the plane of the forces,
must be equal to zero.
Remark. When there is in the plane of the forces a fixed
point, and the forces are in equilibrium, the pressure on the fixed
point = 'tF, which is the same as if every force were transposed
to that point, without altering the direction in which it acts.
32 NON-PAKALLEL FORCES IN A PLANE.
64. Any number of forces act upon a rigid body in one plane,
at different points of the body and in directions not necessarily
parallel; to fnd their resultant.
Let-i?;, F^,...F„loe the forces, and in the plane in which
they act, from a point 0, arbitrarily chosen, draw any two lines
Ox, Oy at right angles to each other. To these lines as
co-ordinate axes refer the given forces and their points of
application (fig 15).
Let «!, a^, ... a„ be the inclinations of the lines in which
the forces act to Ox; x^y^, x^^, ... x^y^ the co-ordinates
of the points of application of the forces; if P be that of
jfj, then a!j= OM, y^=PM. From the point draw OQ
perpendicular to F^ P; and at apply two opposite forces
F', F" each equal and parallel to F^. By this means we have
a force F' acting at 0, and a couple {F^F",) whose moment is
equal ^la -F^.OQ. Or we may say that the force F^ may be
transposed to without altering its direction, if at the same time
we also apply to the body a couple whose moment
= -F,.OQ
= - j; . {ON- QN), since MNis parallel to PQ,
= -F^. {x^ cos HON- y^ sin MPQ)
= — F^. (iCj sin ttj — y^ cos aj
= {F^ cos aj . y^ - {F^ sin aj . x^ ;
or =X^y^- Y,x^,
if we put X, Fj for the components of F^ parallel to the
co-ordinate axes Ox, Oy.
The same method being applied in succession to each one
of the remaining forces of the system, we shall have transposed
all the forces to 0, each preserving its original direction; but
there will be acting on the body besides them a number of couples
in one plane whose moments are
^1^1 - i^i^^i. -3;^,- Y^x^ ... X„y„ - F„a;„.
NON-PARALLEL FORCES IN A PLANE. 33
If Q be the resultant of the couples, and R the resultant of the
forces at 0, we shall have
= t[Xy-Yx)...Kxi.b%
and I^=i^.F cos a)" +{t.F sin a)^ . . Art, 31,
= {t.Xf+{t.Yf
% Y
and tan0 = =-^,
6 being the inclination of the line in which B acts to Ox.
65. The result just obtained is perfectly general, but it
can be simplified, being reducible to a single, resultant, except
when R = Q, i. e. except when SX= and X Y= 0.
For, (1) WhenSX=0 and SY=0, there is no resultant
force acting at 0, and the only resultant is the couple whose
moment = {Xy — Yx).
(2) When the two equations SX= 0, and X Y= are not
both satisfied, change the couple whose moment is 2 {Xt/ — Yx)
into an equivalent couple which has each of. its forces B' It"
equal to B, and place it so that one end of its arm OK (fig. 16)
shall be at 0, and one of its forces {B") exactly opposite to B.
R and B" balance each other and may be removed ; and there
remains only the force J?' acting at the point T such that
OT.CQ^e=OK;
.: B. OT.cose = B. OK,
or, since ^ cos ^ = tX (Art. 31), and B . 0K= t [Xy - Yx),
t{Xy-Yx)
^^~ tx •
Consequently when the two equations 'tX=(i, 2F=0are
not both satisfied, the resultant is a single force B acting at the
point just found, or at any point in the line KB'.
Eemark. From this it appears that when non-parallel
forces, acting in one plane on a rigid body, admit of a single
E. s. 5
34 NON-PAEALLEL FORCES IN A PLANE.
resultant, there is a certain line to any point of which all the
forces admit of heing transposed (each force retaining its original
direction) without their effect being in any respect altered. This
line is EK, and we shall shew in the following Article how. its
equation may be found,
66. When the forces of Art. 64 are reducible to a single
resultant force, to find the equation of the line in which it acts. •
Let x' y be the co-ordinates of any point in the line UK
(fig. 16) in which the resultant acts. Then because this line
passes through the point T and, being parallel to OR, makes an
angle Q with the axis Ox, its equation is
y'-OT=t^-a.e.x,
, t{Xy-Yx) _tY ,
^^ y ~ tx ~sz"'^'
or y'tX-x'tY=t{Xy - Yx).
67. Any number of forces act on a rigid body in one plane at
different points of the body; to find the conditions that they may
balance each other.
Let the system of forces be that of Art. 64, then we have to
consider the two cases of Art. 65. In the second of these cases
the resultant is the force R' (= R) acting at T; there cannot be
equilibrium unless this force vanish, or ^ = 0. But if this be
the case, the second case coincides with the first ; and the re-
sultant is a couple whose moment = % {Xy — Yx)'] there cannot
be equilibrium unless this couple also vanish. Consequently the
conditions of equilibrium are
XX=0, tY=0, S(Xy-Fa;)=0.
These three conditions are both necessary and sufficient.
By referring to Art. 64 we perceive that X,y^ — F,a5, is equal
to the moment of i^, about the point 0, consequently ;S (Xy— Yx)
is equal to the sum of the moments of all the forces about 0.
If then we remember that the point 0, and the directions of the
axes Ox, Oy, were arbitrarily chosen in the plane of the forces,
we may enunciate the conditions of equilibrium as follows :
NON-PARALLEL FORCES IN A PLANE. 35
The algebraic sums of the components of the forces parallel
to any two lines at right angles to each other in the plane of the
forces must he each equal to zero ; and the sum of the moments
of all the forces about any point in the plane of the forces, or
about any axis at right angles to the plane of the forces^ must also
be equal to zero.
68. Suppose that there is in the plane of the forces a fixed
point, or perpendicular to the plane of the forces a fixed axis ; to
find the conditions of equilibrium.
Let the fixed point, or the point where the fixed axis cuts
the plane of the forces, be taken for the point in the investi-
gation of Art. 64. Then the force R which acts at can pro-
duce no effect since it acts on an immoveable point, it is not
necessary then that R should "he equal to zero. But the couple
whose moment is % i^y ~ ^a;), if it exist, will turn the body
about 0, and therefore that there may be equilibrium it is neces-
sary and sufficient that
t{Xy-Yx)=0.
There is therefore in this case only one necessary condition
of equilibrium, viz ; —
TTiat the sum of the moments of all the forces about the fixed
point or axis should be equal to zero.
Eemaek. When there is equilibrium, that is, when the
above condition is satisfied, the pressure on the fixed point is
due entirely to the force R which acts directly upon it. Hence
the pressure on the fixed point is the same as if all the forces
which act on the body were transposed to the fixed point with-
out altering their directions.
CHAPTEE IV.
on forces, not in one plane, which act upon
d;iffbebnt points of a rigid body.
69. If the directions of the forces all pass through a point
we may transfer them to that point, and find their resultant by
Chapter I. or II.
70. In the present Chapter we shall meet with couples of
which the planes are not parallel. We can however always
reduce them to other couples in the planes of rectangular co-
ordinates. It is necessary therefore only to observe that when
a couple acts in a co-ordinate plane, it wiU be considered a posi-
tive couple when its axis stands on the positive side of that
plane. Thus a positive couple
in the plane yz has its axis coinciding with + Ox,
xz + Oy,
xy + Oz.
71. Parallel forces not in one plane act, on different points of
a rigid hody; to find their resultant.
Take any point (fig. 17) in the rigid body, fi-om which
draw Oz parallel to the direction of the proposed forces, which
take for axis of z. Draw Ox, Oy in any directions at right
angles to each other and to Oz, and take them for the axes of x
and y. Let Z^, Z^...Z^ be the forces ; P the point where the
line in which Z^ acts cuts the plane xy. OM=x^, MP=y^ the
co-ordinates of P. Complete the parallelogram OMPN and join
OP. At the point apply two pairs of opposite forces Z', Z"
each equal and parallel to Z^ ; these do not afiect the system.
PARALLEL FORCES NOT IN A PLANE. 37
Now it has been shewn in Art. 55, that when two equal
parallel forces act in the same direction at the extremities of one
of the diagonals of a parallelogram, they may be transposed to
the extremities of the other diagonal. Let us on this principle
transpose Z^ to M, and Z' to N. We have then, one, force Z'
acting at in the direction Oa, and two couples in the plane
xz, yz whose arms are OM, ON, the former couple being nega-
tive. By this means we have transposed the force Z^ to 0,
retaining its proper direction, and have introduced the couples
— Z^x^, +Z^yi in the planes xz, yz respectively. Proceeding in
the same manner with the remaining forces ^, ^...^, we shall
have, instead of the original system, the forces Z^, Z^,...Z^
acting in the line Os, which (by Art. 23) have a resultant
R^t.Z.:. (1);
and, in the plane xz, a set of negative couples, which (by Art. 52)
are equivalent to a single couple in that plane whose moment
=^-t{Zx);
and, in the plane yz, a set of positive couples which are equi-
valent to one whose moment
= t{Zy).
If C be the moment of the resultant of these two couples,
and 9 the angle which its arm makes with Ox, we shall have
from Art. 55,
G.cose = t{Zx), and O . %va.e = t {Zy) ;
.-. CP={t.Zxy+{t.ZyY (2),
andtan^ = i-^ (3),
Equation (1) gives the resultant force acting parallel to the
original forces at the origin of co-ordinates ; and equations (2)
(3) give the magnitude of the resultant couple, and the position
of its plane.
72. We have determined the position of the arm of the
resultant couple. That result supposes, as in fig. 17, a negative
force acting at that extremity of the arm which is at 0, and a
38 PARALLEL FORCES NOT IN A PLANE.
positiye force at the other extremity. It will sometimes be more
convenient to know the position of the positive axis of the
couple. Let a be the inclinations of this axis to the axis of x.
Then
t.Zy , . t.Zx
cos a = — Yr~ > ^^^ sm a = ^y— .
In which equations G is to be accounted positive.
73. The results obtained in Art. 71 are perfectly general,
but they admit of reduction to a single force except when M or
%Z=0.
(1) When 'tZ= 0, there is no force acting at 0, and the
only resultant is the couple whose moment is G.
(2) When "tZ is not equal to zero, change the couple whose
moment is G into an equivalent couple which has each of its
forces B'B" equal to i? or %Z; its arm will be equal to ,-^;
place this couple as in fig.. 18, so that one of its forces B"
balances the resultant B. By this mode the whole are reduced
to a single force B' {=%Z) acting at a point P whose co-ordinates
x'y are known from the equation
x' = OPcos e, and y' = OP sin 6
G „ G . a
t.Zx _ t.Zy
~ %Z ~^ZZ~'
These equations are free from ambiguity.
74. To find the conditions that the system of forces in Art. 71
may balance each other.
We must consider the two cases mentioned in the last Arti-
cle. In the second of these cases the resultant is the force "ZZ
acting at a point whose co-ordinates are
S. Z x , % .\Z/
PARALLEL FOllCES NOT IN A PLANE. 39
There can be no equilibrium as long as this force exists ;
we must therefore have %Z= 0. But if this be the case the
2nd case coincides with the first; so that the resultant is a
couple whose moment is G. There cannot be equilibrium there-
fore unless (r = ; an equation which is equivalent to S . ^a; =
and 'Z .Zy = Q. Hence the conditions of equilibrium are
tZ=0, %.Zx = 0, %.Zy = 0;
which three conditions are both necessary and sufficient.
75. General definition of "the moment of a force about
a line^
If the direction of the force be perpendicular to the given
line, the moment is equal to the product of the force into the
length of a line which is perpendicular hoth to the line in which
the force acts and the line about which the moment is required.
If the direction of the force be not perpendicular to the given
line, it must be resolved into two components, one perpendicular _
and the other parallel to the given line; the moment of the
former will be found by the definition just given, and that of the
latter will be zero.
76. According to this definition Z^y^ and Z^x^ are the
moments of Z^ about the axes of x and y respectively; and
hence we may state the three conditions of equilibrium of parallel
forces acting on a rigid body as follows :
The sum of all the forces must he equal to zero ; and the
sums of their moments about any two lines at right angles to each
other in a plane which is perpendicular to the direction of the forces
must he respectively equal to zero.
77. To find the conditions of equilibrium of the forces in
Art. 71, when there is in the hody a fixed point; or a fixed line
at right angles to the direction of the forces.
(1) When there is a fixed point.
Let it be taken for the point in Art. 71 ; then, as by
this an-angement R acts upon an immoveable poiijt, it is not
40 PARALLEL FORCES NOT IN A PLANE.
necessary that %Z should be = ; but as the couple O would
turn the body round it is necessary and sufficient for equili-
brium that G — Q, or that
%{Zx)=Q, and 2(%)=0.
That is ; the sums of the moments of the forces about any
two lines drawn from the fixed ^oint at right angles to each other,
in a ^lane perpendicular to the direction of the forces, must be
separafeli/ equal to zero. ^
Remark. In this case, i. e. when there is equilibrium, the
pressure on the fixed point is S^ acting directly upon it;
i. e. it is the same as if all the forces were transposed to that
point without altering the direction in which they act.
(2) When there is a fixed line in the body, at right angles
to the direction in which the forces act.
Let it be taken for the axis Oy in Art. 71, being any
point in it.
Then since the force B acts upon a fixed point it is not
necessary for equilibrium that it should be = 0. Also the
couple O is equivalent to the two S {Zx), ^ {Zy) : the latter
of which being in the plane yz can be so placed that its forces
shall both act upon points in the line Oy, which being immove-
able, it Is not necessary that this couple should be equal to zero.
The remaining couple % (Zx) tends to turn the body about the
^ fixed line Oy, so that there cannot be equilibrium as long as it
exists. Wherefore the only condition which in this case is
necessary and sufficient for equilibrium is
t{Zx) = 0,
that is, the sum of the moments of all the forces about the fixed
line must be equal to zero.
Eemark. In this case the pressure on the fixed axis is
equivalent to the force XZ at 0, and the couple S {Zy) ; as in
Art. 59 (2) these are equivalent to a single force "tiZ acting at
a point K in Oy such that
0K=
%z
RESULTANT OF THREE COUPLES. 41
The force %Z acting at this point represents the pressure on the
axis.
But the pressure may be otherwise represented, for a com-
parison of the equation 0K==-^1 with the result found in
Art. 59 shews that the pressure on the axis is just the same as if
the forces Z^Z^Z^ . . . were transposed to the fixed axis and applied
without changing their directions to points in the axis at the
respective distances i/^i/^, ... from 0: that is, if through every
force we draw planes at right angles to the fixed axis we may
transpose each force without altering its direction to the point
where the corresponding plane cuts the axis. The forces thus
transposed produce the same pressure on the axis as the given
system.
78. To find the resultant of three couples which act upon a
rigid hody in different planes, no two of which are parallel.
From any point (fig. 6) draw three lines OA, OB, OG to
represent the axes of the couples, the moments of which are L,
M, N; complete the parallelepiped, and join OD, OF. Then
the couple whose axis is OD is equivalent to L, M whose axes
are OA, OB: and OE is the axis of a couple which is equi-
valent to 00, OD; i. e. to the three couples L, M, K
79. To find the resultant of three couples L, M, N whose
planes are mutually at right angles.
From any point (fig. 7) take QL, OM, ON to represent
the axes of the given couples. Then as before we may shew
that OP represents the axis of the resultant couple O. Let a,
j8, 7 be the angles POL, POM, POiV between the axis of G and
the axes of L, M, N. Then since
OL = OPcos a, 0M= OP cos ^, 0N= OP cos 7 ;
.•.L=G cos a, M=G cos /3, N=G cosy,
and .'.L' + M''+N'=G\
From which four equations the magnitude and position of the
resultant couple are known.
E. s. 6
42 ANY FORCES ACTING ON A EIGID BODT.
80. By means of the four equations just given we may-
resolve a couple into three components acting in planes at right
angles to ei>ch other.
81. To find the resultants of any forces, acting on different
points of a rigid body, in lines which are neither parallel nor in
one plane.
Take any point (fig. 17) of the rigid hody as origin, and
from it draw any three lines perpendicular to each other for axes
of co-ordinates.
Let x^y^s^, x^y^s^, ...be the co-ordinates of the points- at
which the forces act; resolve each force into three components
parallel to Ox, Oy, Oz.
Denote the components parallel to a; by X^, X^, X^...
2^ by Y„ Y„ F,...
■ « by Z„ Z„ Z,...
The resultants of the last set of forces we have already found
(in Art. 71) to be
a force %Z acting at in the line Oz,
a couple S {Zy) acting in the plane yz,
and a couple — S {Zx) acting in the plane xz.
The forces FjFgFj ...form a system of parallel forces, of
which the resultants may be deduced from those of ^^^... by
writing Y, x, z for Z, y, x respectively : they are therefore equi-
valent to
a force % Y acting at in the line Oy,
a couple S ( Yx) acting in the plane xy,
and a couple — % ( Yz) acting in the plane zy.
And in these, writing X, z, y for Y, x, z we find the forqes
Xj, Xjj, Xj ... equivalent to
a force 1X acting at in the line Ox,
a couple % [Xz) acting in the plane zx,
and a couple — 2 {Xy) acting in the plane yx.
ANY FORCES ACTING ON A RIGID BODY. 43
Collecting these results it appears that the original forces
are equivalent to tX, tY, SZ acting at 0; and the three
couples
t{Yx)-t {Xy) =^{Yx-Xy) in the plane xy,
t {Xz) - t (Zx) =t{Xz- Zx) in the plane zx,
and t{Zy)-t{Yz)=t {Zy - Yz) in the plane yz.
Now if R he the resultant of the forces acting at 0, and a, /3,
7 the angles which the line in which it acts makes with Ox, Oy,
Oz ; and if G he the resultant of the couples, and a', /S', 7' the
angles which its axis makes with Ox, Oy, Oz, we have hy Arts.
34, 79,
i?cosa = SX, Bcos^ = tY, Ecoay = tZ,
R={tXy+{tY)' + {tZ)':
and G cos a' = S {Zy — Yz) = L suppose
aco5p' = t{Xz-^Zx)=M
Gcosy' = %{Yx-Xy) = N
G^ = L^ + M' + Jsr.
These eight equations give both the magnitude and direction
of the resultant force which acts at the origin of co-ordinates ;
and the magnitude and position of the axis of the resultant
couple. These results are quite general, but we shall now shew
that under certain conditions the original forces admit of a single
resultant.
82. To jind the condition that the forces in Art. 81 may
admit of a single resultant, and to find the magnitude and position
of it.
If (? be = 0, no reduction is necessary ; but if not, change
the couple G into an equivalent couple, whose forces R'R" are
each equal to R; place this couple so that one of its forces
(as R') shall act at 0, and if possible in a direction opposite to
R ; in this case R and R' balance each other and may be re-
moved ; there will then be left only the force R", which is the
same as if the force R had been transposed to R", and the couple
44 ANY FORCES ACTING ON A EIGID BODY.
taken away. It appears, then, that a couple and a force are
reduced to a single force (where the problem is possible) by '
taking away the couple and transposing the force to some other
point. The possibility of being able to do this, depends on its
being possible to place the forces R and E in the same line.
The student will perceive that this can be done only when the
force B acts in a line which is perpendicular to the axis of the
couple Q, the analytical condition of which is
= cos a cos a' + cos /3 cos j8' + cos 7 cos 7'
_tX L tYM tZN
~ R -a'^ B 'G^ B •(?•
Hence the conditions required are
(1) if must not be = ;
and (2) LtX-ir Mt Y+ NtZ must = 0.
We have yet to find the line in which the resultant force B"
acts.
We have remarked above, that B" is the force B transposed
without altering its magnitude or direction. If we had begun
the investigations of Art. 81 by taking a point in B" for the
origin of co-ordinates, we should have found B" acting at that
origin, and no resultant couple ; that is, denoting the co-ordi-
nates referred to this origin by x", y", z" we should have found
= S (X»" - Zsl'),
and = S(ra;"-Z/);
these are in fact the conditions that the origin may be a point in
the single resultant force. Let a?', y, z'- be the co-ordinates of
this origin referred to the original origin, x, y, z being the same
as before ; then x'= x - x', y" ~y- y', z" = z- z', which being
substituted in the above equations give
ANY FOllCES ACTING ON A RIGID BODY. 45
y'%Z- z't Y=t{Zy- Yz) == L,
z'tX- x'tZ = t{Xz~ Zx) = M,
x'tY-y'tX=t{Yx-Xy)=N,
x, y', z are the co-ordinates of any point in the line in which
i2" acts. There being three, equations between these quantities,
it would seem as if there existed only a single point at which
^".can be applied, which is contrary to ^rt. 21: but if we
multiply these equations by %X, % Y, "ZZ and add the results
we shall find
= LtX + MtY+NtZ,
which being satisfied by hypothesis, the three equations are not
independent, but any one is derivable from the other two. Con-
sequently any two of these are the equations of the line in which
the single resultant acts.
83. When the farces in Art. 81 do not admit of being reduced
to a single force, they can be reduced to a force and a couple the
axis of which is parallel to the force.
For let <^ be the angle between the axis of the couple G
in Art. 81, and the force B. Kesolve G into two components
G cos ^, G sin ^ whose axes are respectively parallel and per-
pendicular to R. The latter of these, being compounded with B
as in the last Article, will be destroyed, ■ and R will be trans-
posed to some other point of the rigid body, without altering
its direction ; it is therefore still parallel to the axis of the couple
whose moment
= G cos <\)
= G (cos a. cos a' -H cos /3 cos /3' + cos 7 cos 7')
LtX + MtY+NtZ
B
This appears to be the simplest form to which the forces in
Art, 81 are in general reducible. They may however be pre-
sented in another simple form as in the following Article.
84. The forces in Art. 81 can he reduced to two forces acting
in two Unes which in general do not meet; and to find the shortest
distance between these lines,
46 ANY FORCES ACTING ON A EIGID BODY.
Let them be reduced as in the last Article to a force E and
the couple Gf cos (f).
Let Q (fig. 19) be the point at which R acts ; and let the
couple G CO? be placed so that one of its forces K acts at Q,
PQ being its arm. Then QR being parallel to the axis of the
couple is perpendicular to QK; hence li H\>& the resultant of R
and K, and -4^ be the angle HQK, the forces are now reduced to
K at P, and ^at Q, such that
„ „ (tCOS^
H sin yjr = R ;
, , , R.PQ
and tan vr = -79 ; .
^ - 6r cos 9
Now PQ being at right angles both to QH and Pff" is the
minimum distance between them. It appears from the above
equations that PQ is arbitrary ; but when it is of given length
then both K and JI are known, and their relative position from
the last- equation. Q is known by the preceding Article.
85. To find the equations of the line in which E, acts, and of
the plane in which G acts, in Art. 81.
Since R passes through the origin its equations are
t t I
X y ^ z
cos a cos /3 cos 7 '
^ _ y' _ 2'
Again, we may suppose the plane of G to pass through the
origin. And since a', /3', 7' are the angles which a perpendicular
upon it makes with the co-ordinate axes, its equation is
X cos a' + y' cos /8' + «' cos 7=0;
ANY POECES ACTING ON A EIGID BODY. 47
.-. Lx +My' + Nz'=(i
is the equation to the plane in which G acts.
86. The conditions that the plane of O may be perpen-
dicular to the line in which R acts uic ^r^= =rT^==r^.
87. To find the equations of the line in which the resultant
force acts when the resultant couple acts in a plane at right angles
to it. (Art. 83).
Let 0' he any point in the line in which the resultant acts in
this case ; x, y, z' its co-ordinates referred to the origin used
in Art. 81. If with the origin 0' we were to proceed as in Art.
81, we should find a resultant R acting at 0', and a couple G',
the plane of which would he found to be perpendicular to the
direction of R ; and therefore
L' M' N'
where L', M', N' represent the quantities
t {Zy" - Yz"), X iXz" - Zx"), t ( Yx" - Xy"),
and x", y", z" are the co-ordinates of a point referred to the
origin 0': hence x" = x — x', y" = y—y', z" = z — z', as in
Art. 82 ;
t{Z{y-y')-Y{z-z')} _ t{X{z-z')-Z{x-x')}
'• XT' 2F
t\Y{x-d)-X{y-y')]
- %Z '
or bringing x, y', z' outside of the symbol S, and writing L, M,
Niox their equals, the equations required are
z
,tY ,%Z^ L _,tZ ,tX M_
%x~y %x^^X~ sr"'' XY'^tY
,%X .tY N
48 ANT FOECES ACTING ON A RIGID BODY.
88. In Art. 83 we were able, by transposing B to destroy
the couple G sin ^. If afterwards we transpose B to some other
point, we shall thereby introduce a new couple, the axis of which
being at right angles to the axis of the couple G cos <j> would be
compounded with it, and make a resultant couple greater than
G cos ^. ' Hence to whatever point B be transposed, the resultant
couple will always be greater than in Art. 83. Consequently
the resultant couple is a minimum when its axis is parallel
to the resultant force. This is sometimes called the principal
couple.
89. Def. The line in which B acts when the resultant
couple is a minimum, is called the central axis. Its equations
are found above in Art. 87.
90. If B be transposed from the central axis to a distance a
from it, a couple is thereby introduced whose arm is a and
moment Ba ; consequently the resultant couple for this position
of B is ViiV + G^ cos^ 0, which is constant as long as a is con-
stant. Hence if we construct a cylindrical surface having the
central axis for its axis, the surface of this cylinder will be the
locus of the origins, which will give equal resultant couples.
91. To find the conditions that the forces in Art. 81 inay
balance each other when the iody upon which they act is free.
(1) Suppose the direction of B to be not parallel to the
plane in which G acts ; then since E and G cannot in this
case be reduced to a single force, they must be separately equal
to zero ;
which are equivalent to the six following :
= tX, O^tY, = -ZZ,
= L, = M, = K
ANY FORCES ACTING ON A RIGID BODY. 49
(2) Suppose the direction of R to te parallel to the plane
in which Q acts; then R and G can be reduced to a single
force, the effect of which reduction is to transpose R and de-
stroy G. There can therefore he no equilibrium unless R = 0;
it is necessary therefore that R should be = 0. But if J? = 0,
R and G cannot be reduced to a single force ; that is, G cannot
be destroyed by transposing iJ; it is therefore also necessary
that G should separately be = 0. Hence the conditions of equi-
librium are the same in this as the preceding case; and, ob-
serving that i, M, N are the moments of the forces about the
lines Ox, Oy, Oz, we may thus state them in words :
The sums of the resolved parts of the forces parallel to any
three lines at right angles to each other must he separately equal
to zero ; and the sums of the moments of the forces, about any
three lines at right angles to each other and passing through a
point, must be separately equal to zero.
92. To find the conditions that the forces in Art. 81 may
balance each other when one point of the rigid body is fixed.
Let this point be taken for the point in Art. 81. Then
since by this arrangement R acts upon a fixed point, it is not
necessary for equilibrium that R should vanish ; but as the
couple G would turn the body about this point, it is necessary
and sufficient for equilibrium that G' be = ; that is, that
L = 0, M= 0, N= 0.
Or, in words : The sums of the moments of the forces, about
any three lines at right angles to each other passing through the
fiooed point, must be separately equal to zero.
Eemark. The pressure on the fixed point is represented
by R acting directly upon it : i. e. it is the same as if all the
forces of the system were transposed to the fixed point without
changing their directions.
93. To find the conditions that the forces in Art. 81 may
balance each other vjhen there is in the body a fixed axis.
E. s. 7
50 ANT FORCES ACTING ON A RIGID BODY.
Let the fixed axis be taken as the axis of z in Art. 81, and
any point in it as the point ; then since B, acts upon a fixed
line it is not necessary for equilibrium that B, should be equal
to zero ; also the couples L, M, acting in the planes yz, xz, can
be turned round and so placed that their forces shall all act upon
the fixed line Oz; but the couple N acting in the plane xy
cannot be so placed, a&d therefore as long as it exists it will
turn the body round the line Oz ; consequently it is necessary
and suflScient for equilibrium that N= ; — or in words,.
' The sum of the moments of the forces dbawt the fixed axis must
he equal to zero.
Remark. The pressure on the fixed axis is represented
by the force JB at the origin and the forces of the two couples
L, M applied directly to the axis. But R is equivalent to the
three 2X, "ZY, "tZ; of which %X can be compounded with
the couple M (the forces of which are in the same plane with
it) as in Art. 59 (2) ; the result of this compounding is a single
force %X acting at a point in the axis the abscissa of which is
yY . The force S Y may in like manner be compounded with
the couple M; and the result in this case is S F acting at the
M
point — =-y.. Hence then the pressures on the axis are repre-
sented by
{^),
"ZX fat the point g^j parallel to x,
ZyI ~Yy) P^'*^^®^ *° ^>
ZZ (at any point of the axis) parallel to z.
The last of these {tZ) may be compounded with either of
the others ; and thus in the most general case the pressure on a
fixed axis may be represented by two forces.
COE. The pressure X^ urges the axis in the direction of its
length, the other two pressures tX, %Y can be reduced to a
ANT FORCES ACTING ON A EIGID BODY. 51
force and a couple, the plane of the couple being perpendicular
to the direction of the force. As this reduction is useful in cer-
tain cases, we shall shew how it may he effected.
Through the fixed axis draw a plane so inclined to the two
forces "tX, 't Y that the resolved parts of %X, S Y along this
plane may be equal and in contrary directions. Let a normal
to this plane make an angle 6 with %X, and therefore an angle
90° — 6 with X Y; then the components
^ f "ZX. cos parallel to the normal,
I "tX . sin along the plane ;
^ ('%Y.sm0 parallel to the normal,
I'tY. cos along the plane ;
of which four components SX.sin^ is equal to 'ZY.cos0 by
hypothesis ; and therefore these two form a couple, and fix the
value of ; for since
XX. sin = tY. cos 0,
%Y
tan0 = ^,
the arm of this couple is
%X'^%Y' ^™™ ^^^'
and therefore its moment is
■.^.■ZX^sm0 + ^.-ZY.cos0
LtY+mx
-{{txy+i%Yy}i'
And the positive axis of this couple is inclined to the axis of x
at the angle 0, given above.
Of the four components mentioned above, the two not yet
reduced are tX.cos0, tY.sin0, acting in one plane on the
points ^. -^- They are therefore (Art. 40 or 59) equi-
valent to a single force
= tX. co30 + t Y. sin = {{tXy + {t Yff,
52 ANY FORCES ACTING ON A KIGID BODY.
acting at an inclination 6 to the axis of x, upon a point in the
■fixed axis the distance of which from the origin is (by Art. 59)
^X.cosg.^-:SF.sing.^ ^ ^^^_^y
SX. cos ^ + Sr. sine I^Xf + (S Yf '
94. To find the, conditions that the forces in Art. 81 may
balance each other, when there is in the body a line moveable
lengthwise but in no other direction.
Let this line he taken as the axis of z in Art. 81, and any
point in it as the point ; then B acts upon this line, and being
resolved into its components XX, ^Y, "ZZ, the first two acting
in directions in which the line cannot move produce no eifect ;
but "ZZ acting in the direction in which the line can move must
be equal to^ zero if there be equilibrium. Also the couples L,
M, being turned round and so placed that their forces shall act
upon the line Oz, produce no effect because they urge it in
directions in which by hypothesis it cannot move: but the
couple N cannot be so placed, and therefore as long as it exists
it -^111 tm-n the body about the line Oz ; it is therefore necessary
that N should be equal to zero. Hence the conditions necessary
and suflScient for equilibrium in this case are,
The sum of the resolved parts of the forces parallel to the given
line must be equal to zero; and the sum of the mament^s of the
forces about the same line must also be equal to zero.
This Art. will be applied when we come to investigate the
power of a screw,
95. The preceding Articles have been enunciated for rigid
bodies only : but since when a flexible body or a body that has
joints is in equilibrium it may be supposed to become rigid
without affecting its equilibrium, all the conditions of equilibrium
before investigated must be satisfied by a flexible body or a body
that has joints. But it is to be noticed particularly that all
these conditions may be fulfilled and yet such a body not be in
equilibrium, for some of its parts may not be in equilibrium.
As a simple instance take the following. A straight rod
placed in a horizontal position with its ends on two props will
THREE FORCES ACTING ON A RIGID BOBY. 53
be in equilibrium ; but a chain, or a rod with a joint in the
middle, so placed would fall. Hence then in equilibrium
flexible and jointed bodies satisfy all the conditions which
rigid bodies satisfy; and besides them such other conditions
as are necessary to secure the equilibrium of every part into
which they are divided by joints : the actions at each joint are,
though unknown generally, equal and opposite upon the two
pai'ts joined there.
96. If three forces acting upon a rigid hody ialance each
other, the lines in which they act must he in one plane, and either
he parallel or pass through a point.
When a rigid body is in equilibrium, we may suppose any
line or point in it to become fixed without affecting the equili-
brium: upon this principle let an axis, not parallel to any of
the forces and intersecting the lines in which two of the given
forces act, become fixed; then these two forces acting upon
fixed points may be removed; which being done the body
having a fixed axis is kept in equilibrium by the remaining
force, which is impossible unless the line in which this force
acts either intersect the fixed axis, or be parallel to it. But it
is not parallel to it by hypothesis, therefore it intersects it. It
appears then that any axis, not parallel to one of the forces,
and intersecting two of them, must meet the directions of all
the forces, consequently they are all in one plane. Again,, since
they are all in one plane they must either be all parallel, or
some two of them must intersect ; in the latter case, the point
of intersection may be supposed to become fixed^ and the cor-
responding forces removed ; and then the rigid body having a
fixed point is kept in equilibrium by a single force, which is
impossible unless its direction pass through tlie fixed point;
consequently, the directions of all the forces either are parallel
or pass through a point.
97. The student will have remarked that when forces (as
in Chaps. I. II.) act on a point, it is not a necessary condition
of equilibrium that their moments about an axis should be
equated to zero. The same is true of every system which is
54 EQUILIBRIUM OF A ETGID BOD'X'.
capable of being reduced to forces acting on a point. Also if
in any case of equilibrium we know that the forces are capable
of being reduced to three forces not parallel, since these by
the last Art. must act in lines passing through a point, the
same is true.
98. In such of the preceding Articles as relate to the
conditions of equilibrium of a rigid body under the action of
a system of forces, the lines parallel to which the forces are to
be resolved, or about which the moments are to be taken, and
equated to zero, have been spoken of as necessarily perpen-
dicular to each other. This necessity, however, has entirely
arisen from the mode in which we have conducted our investi-
gations ; from our having, in fact, assumed the co-ordinate axes
to be rectangular. We shall shew that it may be dispensed
with; and that it is sufficient if the forces be resolved in
directions of, and the moments taken about, any three lines,
provided no two of them are parallel, and all three not in the
same plane. For this purpose it will be necessary to prove
the following propositions.
99. If from a point there he drawn three, lines not in one
plane, and the sums of the components, parallel to them, of all the
forces he separately equal to zero; and also the sums of the
moments of all the forces about them he separately equal to zero;
there will he equilihrium.
For from the proposed point let there be drawn a system
of three rectangular co-ordinate axes Ox, Oy, Oz ; and let one
of the proposed lines make angles ^i, Vu S with them. Then
the sum of the components of the forces in the direction of this
line is
% (Xcos f,) + 1 (FcosT?.) + 1 (^cos rj,
which is equal to
tX . cos fi + S r. cos i7j + tZ. cos §;:
and therefore by hypothesis
= SX.cosfj + SF.cos')7, + S^.cos?,.
EQUILIBRIUM OP A RIGID BODY, 55
Or, if B be tlie resultant of ^X, % Y, %Z; and a, /9, 7 the angles
which its direction makes with the co-ordinate axes ;
= jB cos a . cos ^, + J? cos /S cos ■t]^ + R cos 7 cos Jfj
= R cos a.
Similarly, = J? cos J,
and = i? cos c ;
where a, h, c, are the angles which the direction of i? mates with
the three proposed lines. Now these three equations require
either that "i? should be = 0, or that cos a, cos h, cos c should
each be = ; but this last supposition is impossible, because the
given lines are not all in one plane by hypothesis ;
Again, the couples L, M, N have their axes parallel to Ox,
Oy, Oz respectively : hence resolving them each into two com-
ponents, one of which has its axis parallel to the line ^1, 171, ?i,
and the other has its axisL perpendicular to it, we have the sum
of the former = L cos ^^ + -M"cos rj^ + Ncos, ^, this, being the
couple which tends to turn the body about the line under
consideration, is the moment of all the forces about that line,
and therefore by hypothesis
= i cos ^i + ilf cos T/i + -W cos 5'j
= (? . cos a' cos ^1+ Gr cos /8' cos 1^^+ G cos 7' cos ^^
= Qcosa.
Similarly 0= Gcos h',
and = (r cos c,
a', V, c being the angles which the axis of G the resultant
of L, M, N makes with the three proposed lines. From
these three equations it follows as before, that G = 0; and
we have already shewn that R = 0; consequently there is
equilibrium.
100. CoK. If there be drawn three lines not in one plane,
no two of which are parallel, and the sums of the components,
parallel to them, of all the forces be equal to zero; then the
resultant R is equal to zero. For the first part of the preceding
56 EQUILIBEIUM OF A RIGID BODY.
demonstration applies here, since it does not depend upon the
positions of the lines, but only' on their directions.
101. If ih&re he three lines not in one plane, no two of
which are parallel, and the sums of the components, parallel to
them, of all the forces he separately equal to zero; and if there
he three lines {not necessarily the same as the former) not in one
plane, no two of which are parallel, and the sums of the momsnts
of all the forces about them he separately equal to zero; there will
he equilihrium.
The demonstration contained in the former part of Art. 99,
does not depend at all upon the three lines being drawn from
a point as required in the proposition, and therefore it will
apply here; consequently ^ = 0. From this it follows, that if
our present system of forces be not in equilibrium, their resul-
tant is a couple, Q suppose. Let a', V, c' be the angles which
the axis of G makes with the three, lines mentioned in the
latter part of our proposition ; then resolving G into two com-
ponents, the axis of one being parallel, and that of the other
perpendicular to the first of the three lines, we have the moment
of all the forces about that line (which is equal to the former
component couple)
= G cos a',
which by hypothesis is equal to zero. Hence
= (r cos a'.
Similarly = (? cos h',
and = G* cos c'.
Consequently (? = ; and we have already shewn that
B = Q; therefore there is equilibrium.
CHAPTER V.
ON THE PRINCIPLE OF VIRTUAL VELOCITIES.
102. Def. If the parts of a rigid body, or of a system
of rigid bodies, in equilibrium, be geometrically transferred
through a very small space in any manner, the space moved
over by any particle is called, in Statics, the velocity of that
particle.
The path described by any particle is supposed to be so
small, that it may in every case be taken as a straight line, on
the principle that an arc of a curve ultimately coincides with its
chord.
The velocity of a point, estimated in the direction of the line
in which the force acted upon the point when the body was in
its position of equilibrium, is called the mrtual velocity of the
point.
103. Having given the velocity of a point, to estimate its
velocity in any proposed direction in the plane of motion.
Let AB (fig. 20) be the velocity of a point, -EFthe direction
in which it is required to estimate it. Draw EG perpendicular
to EF; Aa, Bb parallel to EF; and AG parallel to EQ. Then
every line perpendicular to EQ in the plane FEQ is parallel
to, and therefore in the same direction as EF. Hence, to find
the velocity in the direction EF, is the same as to find the space
through which the point has receded from the line EQ. Now
at A the distance from EQ is Aa, and at B the distance is Bh,
consequently the velocity estimated in the direction EF is
E. s. 8
58 PRINCIPLE OP VIRTUAL VELOCITIES.
Bh-Aa = BG=ABcosABG. And ABC is equal to the
angle which the velocity AB makes with the proposed direc-
tion.
Hence we can estimate a velocity in a proposed direction,
by multiplying the velocity into the cosine of the angle at which
it is inclined to the proposed direction.
104. From this it will be seen, that when a particle, which
is acted on by a force, is displaced, the virtual velocity of that
particle will be found as follows; — drop a perpendicular from
the new position of the particle upon the line in which the force
acted before displacement, and the line intercepted between the
foot of this perpendicular and the first position of the point, is
the virtual velocity required. Thus, in fig. 21, let the force F
act upon the point A in the direction AF, and let A be moved
to A'; draw A! a perpendicular to AF, then Aa is the virtual
velocity of A. If A were moved to A" so that FAA" is a right
angle, the virtual velocity of A would be zero. If A were
moved to A" so that the perpendicular A"'a" falls on FA pro-
duced, the virtual velocity Ad" of A is said to be negative.
105. If a rigid hody he displaced in any manner, the velo-
cities of any two of its particles, estimated in the direction of the
line which joins them, are equal.
Let A, B (fig. 22) be two particles of a rigid body, and let
AA', BB' be their velocities. Then, because the body is rigid,
A'B'= AB. Through A draw a plane at right angles to AB,
and upon it drop the perpendiculars A'a, B'h. It will be easily
seen, that the estimated velocity of A is A a; and that of B,
B'h — BA ; and we are to prove these equal. Join a&, and
draw A' parallel to it. The angles at G are right angles, and
therefore
B'b -BA = A'a + B'G-BA
= A'a + A'B' cos A'B' G-BA
= A'a -BA {I- cos A'B'G)
A'B'n
= A'a -2BA. sin' ^^J^.
PRINCIPLE OF VIRTUAL VELOCITIES. 59
But the last term, containing the square of the very
-4 '5'
small quantity sin — - — as a factor, must be omitted in con-
formity with our definition in Art. -102.
Hence Bh-BA=A'a.
This proposition is true, if ^, 5 be two particles of different
bodies connected by a rigid rod, or inextensible string ; for in
the preceding demonstration nothing more is assumed than that
AB' is equal to AB^ which is satisfied in these cases.
106. If the reader should have any doubt respecting the
propriety of omitting the last term, we would recommend him
to reconsider the consequences of the definition in Art. 102,
where it is stated that the displacement of every particle is so
small, that curve lines may be considered as coinciding with
their chords; this requires us to consider the deflection of an
arc from its tangent as evanescent in comparison of the arc
itself, which arc is the velocity with which we J^ave to deal.
Hence
BA{l-cosAB'G),
being the versed sine (or deflection from the tangent) of the
arc which represents a quantity less than the velocity, may be
a fortiori neglected.
107. If the displacements of the two points in Art. 102
be in parallel straight lines through finite spaces, the propo-
sition of Art. 105 will then also be accurately true; and our
definitions in Art. 102, and the property in Art. 103, will also
strictly hold, how large soever be the spaces through which the
particles are displaced,
108. If the particles A, B, in Art. 105, are urged by two
equal forces T, T' in opposite directions along the line AB,
the TOtual velocities hs, hs of A, B for those forces will be
equal, but of contrary signs: and consequently the quantity
Ths + T'hs is equal to zero. Now ii A, Bha two particles of
the same rigid body (or of two different bodies connected in
such a manner by a rod or cord AB that the distance between
60 PEINCIPLE OP VIRTUAL VELOCITIES.
them does not change), their influence upon each other is ex-
erted along the line AB, and is called tension. This tension is
the same for both, but acts upon them in opposite directions,
viz. either to draw them towards each other, or to push them
asunder. Hence it follows, that for the tensions acting between
A and B, TSs + T'Bs =0. The same maybe proved for any
and every two points in a whole system of bodies, provided
they are connected in such a manner, that the distance between
the points of connection is not changed by the displacement.
It is obvious, that the tensions we are now considering, occur in
pairs. Hence it follows, that if the forces of tension throughout
a whole system of bodies in equilibrium be respectively multiplied
by the virtual velocities of the points on which those tensions are
supposed to act, the sum will be equal to zero.
109. If a body rest against a smooth fixed point, there will
be a pressure of the point against the body in the direction of a
normal to the surface of the body. This pressure is one of the
forces which teep the body in equilibrium. Let A (fig. 23) be
the fixed point, FA the surface of the body resting against it,
AB a normal at A, and let the body be displaced without lifting
it off the point, so that A comes to some point A' suppose.
Then the virtual velocity is
^^' cos ^^^' = ^^' sin P^^' ;
but PAA' is an indefinitely small angle, and therefore
AA' sin PAA' is indefinitely smaller than AA', and may be
neglected. Hence, if B be multiplied into its virtual velocity,
the product may be neglected.
110. If a body rest against a smooth fixed curve line or
surface, there will be a pressure of the curve or surface against
the body, in the direction of a normal at the point of contact.
Let PA (fig. 24) be the body resting against the curve line or
surface QA ; and let the body, by sliding and rolling, come into
the position FBA', B being now the point of contact, arid A' the
new position of A. The virtual velocity of .4 = BA' sin ABA';
which, for the same reason as before, may be neglected.
PRINCIPLE OF VIRTUAL VELOCITIES. 61
111. If two smooth bodies of a system rest against each
other, there will be a mutual pressure, which will act upon
them at the point of contact in opposite directions, coinciding
with the common normal at that point. If they are disturbed
without heing separated, the distance between their centres of
curvature, at the point of contact, will remain unchanged ; and,
therefore, the virtual velocities will be exactly equal, but of
contrary signs for the two bodies. If, then, B, E be the equal
pressures exerted by each against the other, and Sr, S/ the
virtual velocities,
Rh- + R'h-'=0.
112. From the last three Articles, it appears that in any
system of bodies kept in equilibrium by the action of external
forces, and by tensions, reactions of smooth fixed obstacles,
and mutual pressures of smooth bodies of the system, the sum
of the products of each tension, reaction, and pressure, into the
corresponding virtual velocity of the point on which it acts, is
equal to zera.
The student will remark, that the Articles referred to, are
only true when the displacement of the system is so small as to
agree with the definition of a velocity given in Art. 102 : also, in
the case of pressures, the surfaces must be smooth, and the con-
tact must not be broken; and in the case of tensions, the
connecting line must be of unaltered length.
113. Let there he any number of connected bodies of a system
kept in equilibrium by the action of external forces, and also by the
tensions of connecting rods, cords, <fcc., by the reactions of smooth
fixed obstacles, and by mutual pressure of smooth parts ; then, if
each external force be multiplied into the virtual velocity of the
point on which it acts, the sum of all such products for the whole
system is equal to zero. It is necessary (as the reader will see
from- the preceding Articles), in geometrically displacing the
system, that no contacts be broken, and that rods and cords
remain of the same length as in the equilibrium position. This is
the principle of virtual velocities.
If a particle of one of the bodies be acted on by the external
forces i^i, F^,...F^ and -by the tensions T^, T^,...T,, and the
62 PEINCIPLE OP VIRTUAL VELOCITIES.
reactions and pressures R^, B^,...B„, we may consider that point
as free, and kept in equilibrium by the action of all these sets of
forces,
Letai/3j7j a^^^%...a^^^% be the angles which the forces
i''j, i^2'"^ ™^^6 with the co-ordinate axes ; afi^c^, apf^...apf^
and a'jS'ic'i, a\h\c\...dJ>'^o'^ similar quantities for T^, T^,... T^
and jB,,5j,...i2„.
Then, because the particle is in equilibrium under the action
of these forces, therefore (Art. 38)
= S (-Fcos a)+t {Teas a)+t{fi cos a'),
= t (i^cos/3) + t-(!rcos b)+%{B cos 6'),
= S (i^cos7) + 1 {Tcos c)+t{B cos c').
Let now the system be displaced, the Telocity of the particle
under consideration being B8^, and ^f the angles which S8j^
makes with the co-ordinate axes. Then, if Bs^ be the virtual '
velocity for the force F^,
jPjSsj = F^ (cos Hj cos f + cos /3, cos j] + cos 7^ cos f ) 88^ .
Similar expressions are true for the other external forces
which act on the point, and therefore
t{FSs) = {t{Fcosa.)eos^+tiFcos^)cosv+'^{Fcosi)cos^}SS^.
Similarly, if Bt^ and Br^ be the virtual velocities corresponding
to 2; and B^,
t (TBt) = {t (Tcosa) cosl+'Z{Tc()sb)cosri+t {Tcosc) cos^\B8,;
and
% {BBr)={t (-Bcosa')cosf +S {Rcosb') cosij+S (-Bcos c) cos ^} B8,.
Hence by adding the last three equations we obtain
t{FBs) + t{TBt) + t{BBr) = (1)
in which the symbol % extends to all the forces, tensions-, and
reactions which act upon the point under consideration, but has
no reference to the other particles of the system.
We may form equations similar to (1) for every other point
in the whole system upon which forces of any kind whatsoever
act. If all these be added together, the terms belonging to the
PRINCIPLE OF VIRTUAL VELOCITIES. 63
tensions along the lines which join points of the same body, and
also those which act along rods and cords connecting two points
of. separate bodies of the system ; and likewise the reactions of
fixed points, and surfaces, and the mutual pressures of two bodies
of the system, will all disappear, by Art. 112,' in forming the
sum. But these, together with the external forces, are all the
forces which act on the system ; consequently, there remains only
the equation
where %' extends to all the points of the systenf upon which
external forces act, S' and % together denote that the sum of the
products of all the external forces which act upon all the points
of the system into their respective virtual velocities is to be
taken, and the equation shews that this sum is equal to zero ;
which is the principle of virtual velocities. It is not necessary
to employ both S and X', if we suppose X to extend over the
whole system, the equation may be written
% (FSs) = 0,
which is called the equation of Virtual Velocities.
114. The great advantage of the equation of virtual veloci-
ties consists in this, that it furnishes at once a relation among
the external forces which act upon a system, free, from tensions
and pressures. Since the bodies are rigid, and supposed to be
connected by strings or rods of unchangeable length, it is
obvious that, in general, when one part is arbitrarily disturbed,
the disturbance of the other parts will depend upon it by geo-
metrical relations. In this case, Ss^ being given, Zs^, Ss, ... will
be determinable in terms of Ss, ; and these values being written
in the equation % (FSs) = 0, will give only one relation among
the forces, and will not therefore enable us to find the forces
themselves, if their number exceed two.
It will, however, sometimes be possible to disturb one part
of the system without affecting other parts ; or the system may
consist of several parts, each one of which it may be possible to
disturb in such a. manner as not to affect the other parts. In
this case it is manifest, that the equation of virtual velocities
64 PRINCIPLE OF VIRTUAL VELOCITIES.
will furnish us as many equations between the forces, as there
are parts of the system which can he independently disturbed.
Now two points can be independently disturbed when no
geometrical relation exists between their virtual velocities.
Wherefore, in using the equation % {FBs) = 0, we must find, from
the geometrical properties of the system, as many of the quan-
tities Ssj, Ssj, Ssj... in terms of the others as possible, and sub-
stitute them in the equation; the virtual velocities which are
stiU left in it are independent, because no geometrical relation
exists among them; and, therefore, the corresponding parts of
the system admit of independent disturbance; we must conse-
quently equate the coefficients of each of these terms to zero.
The resulting equations are the conditions of equilibrium.
To illustrate what is here meant, we will solve the two fol-
lowing problems by the principle of virtual velocities.
115. A particle rests upon a plane curve line, heing acted on.
hy two forces X, Y parallel to the co-ordinate axes : to find the
conditions of equilibrium.
Let y=f{x) be the equation of the curve, x, y being the
co-ordinates of the position of equilibrium of the particle. Then
since after the disturbance the particle still remains upon the
curve, if 2^ + By, and x + hx'he, the co-ordinates of its new posi-
tion they must satisfy the equation of the curve ;
••• y + By =f{xJrhx) =y + d^ .Bx;
.-. Sy = d^y . Sx.
Now Bx, By are the virtual velocities of the particle for the two
forces X, Y;
.*. XBx + YBy = by the principle ;
.■. XBx + Yd^yBx = for all values of Bus,-
and .-. X+ Yd^=0,
which is the condition of equilibrium.
116. A particle rests upon a smooth curve surface acted on
by three forces X, Y, Z parallel to the co-ordinate axes: to find
the conditions of equilibrium.
PRINCIPLE OP VIRTUAL VELOCITIES. 65
Let s =f{so, y) be the equation of the curve surface, x, y, z
being the co-ordinates of the position of equilibrium of the
particle. Then if as + Sas, y^ By, s + Bs be ^he co-ordinates of
the position of the particle after disturbance, Bx, By, Bz are the
virtual velocities of the particle for the forces X, Y, Z respec-
tively ; and therefore by the principle of virtual velocities,
X$,x-\-YBy + ZBz = f).
But because x -f«Sa;, y-\-By, z + Bz are the co-ordinates of a point
in the curve,
z + Bz=f[x + Bse, y + By)
— z + d^z.Bse + dyZ.By;
.". Bz = d^z . Bx + dyZ . By.
By substituting this value of Bx, we have
(X+ Zd,z) Bai+{Y+ Zd,z) By = 0.
There is no geometrical relation existing between By and Bx ;
consequently, the equations of equilibrium are
X+Zd^z = Q, Y+ZdyZ = 0.
117. ^ two forces P, P' whose virtual velocities are Sp, Sp',
act upon a rigid body at different points, and he such that the
equation PSp + P'Sp' = is true for all arhiiyrary displacements of
the hody, then P and P' are equal and act in the same line in
opposite directions.
The equation shews that Bjy and Bp are always zero together.
Now disturb the body in such a way that the point at wTiich P
acts may remain stationary ; then since the body is rigid, the
point on which P' acts must have described a circular arc about
the stationary point ; and as Byp = 0, that arc must be perpen-
dicular to the direction in which P acts, therefore P' acts in the
direction of a normal to the arc, i. e. in a line passing through
the point on which P acts. In the same way it may be shewn
that P acts in a line passing through the point at which P' acts ;
hence they both act in the same line: it will therefore be
possible to disturb the body so that Bp and Bp/ may be equal
E. s. ' 9
66 PRINCIPLE OP VIETUAL VELOCITIES.
in magnitude; and they must have different algebraic signs
(•.■ PSp + P'Sp' = 0), which can only happen, since the body is
rigid, by reason of P and F acting in opposite directions ; and
therefore P and P' are likewise equal.
118. If the eguation % (FSs) = &e true for all arUtrary
displacements of a rigid hody under the action of external forces
F, , Fjj . . . there is equilibrium.
For if not, there will be at most two restfttants (Art. 84) ;
apply forces P, P' equal to these resultants and in the contrary
directions to them, and then the body is in equilibrium under
the action of the forces F^, F^ ... P, P; consequently by the
Principle of Virtual Velocities,
S(PSs)+PSp + PS/ = 0.
But % {FSs) = by hypothesis, and therefore PBp + P'Bp' = :
and hence it follows from the last article that P and P are
equal and act in opposite directions ; consequently they destroy
each other ; they may therefore be removed without affecting the
equilibrium ; hence the body is in equilibrium when Pj, Pj, P^ ...
are the only external forces which act on the body.
119. When a system of connected bodies is in equilibrium
under the action of external forces, pressures, &c., the equilibrium
would not be affected if the connecting joints, cords, &c. were
all to become rigid : and hence any force may be transmitted to
any point of the system in the line of its action (Art. 21), pro-
vided the original point and the new point of application are
not situated in independent parts of the system.
120. If the equation t (FSs) =0 be true for all arbitrary
displacements of a system of connected rigid bodies, there is
equilibrium.
If the system consist of independent parts, let one of those
parts alone be displaced, then for that part % {FBs) = by hypo-
thesis. If that part is not in equilibrium we may apply forces
to each body of it which shall keep each of them in equilibrium :
these forces (Art. 119) may be transmitted and reduced to two.
PRINCIPLE OF VIRTUAL VELOCITIES. 67
P, P, acting upon the part under consideration. Hence reasoning
as in Art. 118, we find Pand P equal and opposite, and therefore
they may be removed without disturbing the equilibrium of the
part. The same may be proved of each of the independent parts ;
and, consequently, the whole system is in equilibrium.
Remark. We have seen that the principle of virtual
velocities is true only when the displacements are so small as
to allow us to consider an arc as coincident with its chord
or tangent. Now the reader who is familiar with the Differential
Calculus will know, that an arc and its tangent coincide ana-
lytically only as far as the second term of Taylor's theorem
inclusive : hence the principle of virtual velocities embraces only
quantities of the first order of smallness. The second term
of Taylor's theorem has been called the differential of the first
term ; wherefore, in applying the principle of virtual velocities,
we ought always to use ds instead of hs. The equation of
virtual velocities in its proper form is % (Fds) = 0. Also because
this equation involves only differentials of the first order, it is
a matter of indifference whether a body rest upon a curve or its
tangent, a surface or its tangent plane ; or on any other curve or
surface having the same tangent or tangent plane at the point
on which it rests.
CHAPTER VI.
ON THE CENTRE OF PARALLEL FORCES, AND ON THE
CENTRE OF GRAVITY.
THE CENTRE OF PARALLEL FORCES.
121. If a rigid hody he acted on at dAfferent 'points hy forces
in parallel directions, there is a certain point through which their
resultant passes, whatever he the position of the hody with respect
to the direction in which the forces act.
Let F^,F,...F„ act on the points A, B ... K (fig. 25) of a
rigid body. From any point in the body draw the rectan-
gular co-ordinate axes Ox, Oy, Oz. Join A, B; and let the
resultant of F^, F^, pass through G. Draw Aa, Bh, Cc parallel
to Os ; join a, b passing through c.
Let aj^yjZj, x^y^z^ ... x„y^z^ be the co-ordinates of the points
on which the forces act; xy'z those of C; and let Q be the in-
clination of AB to ah. Then
^ — z^=Gc — Aa = AG sin Q,
and z^-z' = Bh- Gc=BGwa.e;
z.-z BG F.,, . ^ ,^v
whence we find {F^ +F^ z' = F^z^-\-F^z^.
AgAin, take away the forces jP^, F^ and replace them by their
resultant F^ + F^ acting at G; then if we put x"y"z" for the co-
ordinates of the point through which the resultant of i''^, F^, F^,
or, which is the same, of the two {F^-\- F^ audita passes, we
hare as before
{F, + F, + F,)z'' = {F, + F,)z- + F,z,
= F z +F z +F z .
CENTRE OF PARALLEL FOECES. 69
In this manner, introducing successively one force at a time, until
all have been taken in, and denoting hj x^s the co-ordinates
of the point at which the final resultant acts, we shall at length
obtain,
{F, + F, + F,+ ...+F,)^ = F^z^ + F,z^+ F^z^+ ... + F^z,,
or, more concisely, SF. i = S {Fz).
By similar reasoning we shall obtain
tF.^ = X{Fy),
uni tF.x=t{Fx).
The last three equations determine the values oixy'z; and since
those values do not contain any terms depending on the inclina-
tions (to the co-ordinate axes) of the lines in which the forces
act, those forces may be turned about the points on which they
act without affecting the position of the point whose co-ordinates
are xy^. On this account this point is called the centre of
parallel forces.
122. Def. The product of a force into the distance of the
point on which it acts from a plane, is called the moment of the
force with reject to the plane. Hence % {Fx), ^ [Fy), % {Fz) are
the sums of the moments of the forces with respect to the planes
of yz, xz, xy : and ^F . x, SF. y, %F. z, are the moments of
their resultant with respect to the same planes. Hence, remem-
bering that the co-ordinate planes were taken in any position, it
follows, that the sum of the moments of any parallel forces with'
respect to a plane is equal to the moment of their resultant with
respect to the same plane.
123. If the proposed plane be drawn through the centre of
parallel forces, the moment of the resultant with respect to it will
be zero ; consequently, the sum of the moments of any parallel
farces with respect to any plane passing through their centre is
equal to zero.
124. If %F be equal to zero, there is then no centre of
parallel forces, as we likewise know from Art. 73.
125. The formulae of (121) are true if the co-ordinates are
oblique: and in that case t [Fx), t {Fy), t {Fz) are called the
70 CENTRE OF GRAVITY.
obliqiie moments of the forces with respect to the co-ordinate
planes of yz, zx, xy.
THE CENTRE OF GRAVITY.
126. It has been found by experiment, that under the
exhausted receiver of an air-pump bodies of unequal magnitudes,
and differing altogether in their nature and form (such as a piece
of lead, a shilling, a feather, &c.) fall from the top to the bottom
of the receiver exactly in the same time: from which it has been
inferred, that the Earth exerts an equal force on all equal
portions of matter ; and that the weight of a body at a given
place, measured according to the principles laid down in Arts.
7 — 10, is proportional to the quantity of matter in the body;
that is, if M be the quantity of matter in a body whose weight is
Wat a given place, then
But we have stated in Art. 8, that the weight of a body,
measured by a standard spring, is not the same at all places
of the Earth's surface ; it is in fact (as is shewn in Dynamics)
proportional to the accelerating force of gravity, at the respective
places. This force is generally denoted by g ; and hence . we
have for a given body
W^g.
Consequently, for different bodies at different places WccMg,
For reasons stated in Dynamics we assume that
W=Mg.
127. The size or bulk of a body is called its volume and is
denoted by V: but it is necessary to explain, both with regard to
V and M, that they are expressed in numbers on the following
principle. A known body, composed of matter uniformly dif-
fused through all its parts, is taken as a standard to which all
others are referred. The volume and mass of this body are called
the units of volume and of mass. If a body be V times tlie size,
and contain M times the quantity of matter, of the standard
body ; V and M are taken as the measures of the volume and
mass of that body. Also, supposing the matter of the second
CENTRE OF GRAVITY. 71
body to be uniformly diffused through its parts, if a portion of
it of the same size as the unit of volume contains p times as
much matter, p is called the density of the body; and it is
evident that
M=pV.
128. The direction in which a body descends when let fall
is called the vertical direction ; it may be discovered by suspend-
ing a heavy body by a thread, or by drawing a line perpendicular
to the surface of still water. A plane at right angles to the
vertical is called a horizontal plane; and it is evident, since the
Earth is spherical, that the horizontal plane changes its position
in passing from place to place: but since the distances of the
bodies of systems usually treated of in Statics are exceedingly
small compared with the radius of the Earth (4,000 miles, nearly)
we may consider the sujface of still water as a horizontal plane
to a small extent, and consequently the verticals as parallel.
129. Hence it appears, and from Art. 121, that in every body,
and in every rigid system of bodies, there is a certain point
through which the resultant of the forces which the Earth exerts
on the different parts always passes in every position of the body
or system. This point is called the centre of gravity of the body
or system: it is sometimes also called the centre of mass.
130. One property of the centre of gravity, particularly
worthy of remark, is, that it does not depend at all upon the
intensity of the force of gravity. For divide the whole system
into very small equal molecules, the quantity of matter in each
being m, and their number n, and denote the force exerted upon
a unit of matter by g ; then the force exerted on each molecule
= mg. And if x^y^z^, x^^z^,... be the co-ordinates of the
molecules, and xyz those of the centre of gravity, we have, by
Art. 121,
_ mgx, + mgx^ + mgx^ + . . . to w terms
*~ mg-\-mg + mg + ...ion terms
_ X, + X.;, + X, + .
n
72 CENTRE OF GRAVITY.
Similarly, ^ = ^dJ^i+i^a+inii' ^
andJ = "' + "- + "- +
It appears then, that the co-ordinates of the centre of gravity
are the means* of the co-ordinates of the equal molecules, and
consequently its position is independent of the intensity of
gravity. Hence the centre of gravity of any body is a certain
point within it, the place of which depends only on the relative
disposition of its equal molecules. The investigation of its place
is therefore purely geometrical, and may be applied to any body
whatever; and for this reason we often speak of the centre of
gravity of bodies far removed from the influence of the Earth,
and when, in fact, no reference is intended to be made either to
the Earth or to gravity; the point alluded to being no other
than the one determined from the geometrical principles just
laid down, viz. — that its co-ordinates are the respective means of
the co-ordinates of all the equal molecules of which the body is
composed.
131. When a body is acted on by no other force than
gravity, since the resultant of the forces which act on the particles
of the body passes through its centre of gravity, if that point' be
supported the body will be in equilibrium in every position. For
instead of the forces themselves, we may substitute their resultant,
which will be counteracted by the point of support, and as this
will be the case if the body be turned round that point into any
position whatsoever, it follows that there will be equilibrium in
any position whatever.
132. And since the resultant may be applied at any point
in the line of its direction (Art. 21), if the point of support be
not in the centre of gravity, but in any point of a vertical
passing through it, the body will be in equilibrium. And con-
versely, if a body be suspended from any point in it, it will not
be at rest till the centre of gravity and the point of suspension
are situated in the same vertical.
them.
Hence the centre of gravity of two equal "bodies is ihe middle point between
CENTRE OP GRAVITY. 73
This property may sometimes be employed in finding the
centre of gravity in a practical manner. For if the body be
successively suspended from two points in it, and the correspond-
ing verticals be drawn upon or through the body, their common
point of intersection will be the centre of gravity.
133. It follows at once, from Art. 131, that if all the par-
ticles which are situated in a line passing through the centre of
gravity be supported, the body will rest in equilibrium on that
line in all positions. And the converse is true, viz. — that if a
body rest in equilibrium, in all positions, on a fixed line, the
centre of gravity must be in that line ; for, unless the centre of
gravity were in that line, a position might be found in which
the vertical through the centre of gravity did not pass through a
point of support, and consequently the body would not be in
equilibrium in all positions, which is contrary to the hypothesis.
Hence, if we can find two lines on which a body will rest in
all positions, the centre of gravity will be in their common point
of intersection.
134. Since the resultant of all the forces of gravity, which
act on the particles of a body, may be supposed to act at the
centre of gravity, and is equal to their sum (Art. 121), we may,
in any investigation in which this resultant is required, suppose
the whole mass united at the centre of gravity ; and hence it
becomes important to know the situation of this point in bodies
of different figures.
135. It is not always convenient to divide a proposed body
into equal molecules, as was done in Art. 1 30, it therefore be-
comes necessary, in that case, to use other formulae for the
determination of the centre of gravity.
Let >»,, m^, wij, be very small masses into which the
body may conveniently be supposed to be divided ; a;, y^ z^ ,
x^y^z^, x^y^Zg... their co-ordinates.
Then the forces which urge them are ffm^, gm^, gm^,
respectively; and therefore, substituting in Art. 121, we obtain
E. 8. 10
74 CENTRE OF GEAVITY,
gm, + gm^ + gm,+ ...
m^ + m^ + m^ + ...
_ % imx)
and, similarly,
S (my) _ 2 (mz)
y = '- " — ' —
z ■■
%m ' Xm
136. Since, whatever be the position of the plane yz, we
always have
x.'$m = 'Z (mx),
it appears that the moment, with respect to any plane, of the
whole mass collected at its centre of gravity, is equal to the sum
of the moments of all the molecules, with respect to the same
plane. •
137. If the origin of co-ordinates he in the centre of gravity,
then X {mx) = 0, S {my) = 0, and 2 (wis) = ; for x, y, and z are,
in that case, each equal to zero.
138. Since the mass of a body of uniform density is mea-
sured by the product of its volume into its density (Art. 127) ;
if Pj, Pj, Pj, ... be the densities, and V^, V^, Fg, ... the volumes
of the molecules m,, m^,m^, ... we shall have
w,=p,Fj, m, = p,F„ jw,=/o,F3,...
the molecules being so small, that every part of each one may
be considered of uniform density. Hence, by substitution in the
formulae of Art. 135, we have
P^yt + P.K + PsK+-
_ t{pVx)
CENTRE OP GEAVITY. 75
and «-iMM
S(pF)-
139. If the density of the whole system be the same in
every part, then p^ = p^ = p^ ... and these formulae are simplified
by dividing out p, thus,
™-?iM r,-li]M 7-li]^
«- ^Y ' y- iv ' %v ■
But it is to be carefully observed, that these formula are
only to be applied to such bodies as are of homogeneous mate-
rials.
140. The general application of these formulae depends on
the Integral Calculus, but there are a few cases which can be
made to depend upon the more simple principles of Art. 133,
and with them we shall accordingly commence our series of
examples on the subject of finding the position of the centre of
gravity in bodies of various forms.
All bodies will be supposed homogeneous, or of uniform
density, unless the contrary is mentioned.
141. If through any figure a plane can he drawn, so that the
figure shall be symmetrical with regard to it; that is, so that the
two parts of the figure which are situated on opposite sides of that
plane are perfectly similar and equal; the centre of gravity is in
that plane.
For the volume of the body being similarly disposed on the
two sides of this plane, the moment of the volume on one side is
exactly equal to the moment of that on the other side, with
respect to that plane, and these moments will have contrary
signs, and therefore their sum will be equal to zero. But this
sum (Art, 139) is equal to the moment of the whole volume,
collected at its centre of gravity, with respect to the same plane ;
which cannot be the case unless the centre of gravity be in that
plane.
76 CEBTTEE OF GRAVITY.
142. Hence, if we can find two such planes differently
situated, the centre of gravity will be in the line of their inter-
section ; and if we can find a third plane, the centre of gravity
will be that point where it cuts the line of intersection of the
other two ; in other words, it will be the common point of inter-
section of any three planes, by which the figure can be sym-
metrically divided,
143. It follows, from these properties,-^
(1) That the centre of gravity of a sphere, or of a spheroid,
or of a cube, is its centre.
(2) That the centre of gravity of a parallelopiped is the
middle point of one of its diagonals; and of a cylinder the
middle point of its axis.
(3) That the centre of gravity of any figure of revolution is
some point in the axis.
144. When we speak of the centre of gravity of a line, or
of a plane figure, it is to be understood that the line consists of
material particles, and the plane figure of a single lamina of
particles, or else, that the thickness is everywhere the same, and
inconsiderable.
145. Hence the centre of gravity of a straight line is its
middle point ; of a circle, or, ellipse, or square, its centre ; and it
will follow, from reasoning precisely similar to that of Art. 141,
that if we can draw two straight lines in a plane, by each of
which the figure is divided into two equal and symmetrical
parts, the centre of gravity is the point of their intersection.
This property will enable us to determine at once, by inspection,
the centre of gravity gf almost all regular plane figures.
146. To find the centre, of gravity of a plane triangle,
' Let ABC (fig. 26) be the triangle, bisect one of the sides
as 50 in D, and join AD. Then we may suppose the triangle
made up of material particles, arranged in lines parallel to BC;
let Ic be any one of them. Then, by the similar triangles
BAD, bAd,
BD : DA :: hd : dA,
CENTRE OP GKAVITY. 77
and, similarly, BA : DG :: dA : dc,
.-. BD '. DG :: Id : dc.
But BD =DG, therefore bd = dc; and consequently, <? is the
centre of gravity of ho.
Similarly, the centre of gravity of every other line, parallel
to BC, of which the triangle consists is somewhere in AD ; con-
sequently the whole triangle would rest in equilibrium on AD,
and therefore its centre of gravity is in AD (Art. 133). In the
same manner it would appear that the centre of gravity of the
whole triangle is in BE, which bisects A C, and hence G, the
point of intersection of AD and BE, is the point required.
Join DE, then because GA, CB are divided at E, D in
the same proportion, viz. each bisected, therefore DE is parallel
to AB; and, therefore, the angle DEG- is equal to the angle
ABO, and angle ED G to the angle BAG, and. consequently the
triangles ABG, DEG are similar ;
.-. AG : DG :: AB : DE
:: AG : EG :: 2 : 1.
Hence AG = 2DG,
and .-. AD=AG + DG = 3DG;
.-. DG = ^AD.
147. If three eqtcal bodies have their centres of gravity
situated in the three angular points of a triangle, the centre of
gravity of these bodies will coincide with that of the triangle.
Let A, B, G be the centres of gravity of the three equal
bodies, then BD being equal to DG, the two bodies at B, C will
be in equilibrium on D ; and therefore the three bodies will be
in equilibrium on a line passing through A, D; in the same
manner they will be in equilibrium on BE, and therefore G is
their common centre of gravity.
Hence (Art. 130) the distance of the centre of gravity of a tri-
angle from any plane, is the mean of the distances of its angular
points from the same plane.
78 CENTEE OP GRAVITY.
148. To find the, centre of gravity of a quadrilateral figure.
LetABOD (fig. 27) te the trapezium; AO,BD its diagonals
intersecting in j&; (r its centre of gravity ; draw GI, (^^ paral-
lel to the diagonals. Then, supposing the trapezium to be made
up of the two triangles ADC, ABC, we have (Art. 130),
(trapezium ABCD) . (perpendicular from G on AC)
— {AABC) . (perpendicular from its centre of gravity on AC)
— {AABC) . (perpendicular from its centre of gravity on AC)
= J (AABC) . (perpendicular from B on. AC)
— J (AABC) . (perpendicular from B upon A C).
Now the triangles ABC, ABC, having a common base AC,
are proportional to the perpendiculars from B and D on AC,
which are also proportional to BE, BE respectively ; hence, in
the above equation, instead of the triangles ABC, ABC, and
the trapezium, which is their sum, write respectively the quanti-
ties BE, BE, and BE+BE, to which they are proportional;
and, instead of the perpendiculars from B, B and G, or I, which
is equal to it, write respectively BE, BE, and EI, which are pro-
portional to them ; and then we have
{BE + BE) . EI= IBE'- ^BE^
= J {BE+ BE) {BE- BE) ;
.-. EI=i{BB-BE).
And, similarly, EK=^{AE-CE).
Hence, setting off EI equal to one-third of the excess of EB
above ED ; and EK equal to one-third of the excess of AE
above EC; and drawing IG, KG parallel to the diagonals of
the trapezium, G will be the point required.
149. To find the centre of gravity of any other rectilinear
figure we must divide it into triangles, and suppose each triangle
collected at its own centre of gravity; we can then find the
common centre of gravity of the whole by the formulse of
Art. 139.
CENTRE OP GRAVITY. 79
150. To find the centre of gravity of a triangular pyramid.
Let A (fig. 28) be the vertex, and BOE the base of the
pyramid. E, H the centres of gravity of the base and the face
A CD. Join AE, BE, BE, AH. Then, because E is the centre
of gravity of the base, therefore BE produced, bisects CD. For
a similar reason, AH produced, bisects GD ; and therefore BE,
AH intersect in F; consequently, AE, BH, which are in the
plane ABF, intersect each other in some point O.
Now we may suppose the pyramid made up of triangular
laminae of particles, situated in planes parallel to the base ; let
cbd be one of them, cutting ^i''in/, and AE in e. This triangle
is, of course, exactly similar to the base of the pyramid, and
being parallel to it, cd must be parallel to GD ; and therefore
the triangles GAF, cAfaxe, similar,
.-. cf : Af :: GF : AF;
for a similar reason, Af : df :: AF : DF;
.: cf : df :: CF : DF;
but GF being equal to DF, cf must be equal to df, and con-
sequently the centre of gravity of the triangle chd must be in the
line bf. Again, AFB being cut by parallel planes, fb must be
parallel to FB, and the triangles FAE, fAe are similar,
.-. fe : Ae :: FE : AE;
but, for a similar reason,
Ae : be :: AE : BE,
.-. fe : be :: FE : BE.
But BE=2FE, and therefore be= Ife, consequently e is the
centre of gravity of the lamina bed. In the same manner it
may be proved that all the triangular lamins^ of which we have
supposed the pyramid to consist have their centres of gravity
in AE, wherefore the pyramid would balance on AE in all
positions ; and, consequently, the centre of gravity is in that
line. For like reasons, it is in the line BH, and therefore G,
the point of intersection of AE and BH, is the centre of gravity
of the pyramid.
80 CENTBE OP GRAVITY.
Join HE. Then, because FE : FB :: I : Z :: FH : FA,
therefore HE is parallel to AB, consequently the triangles HEG,
BA G are similar ;
.-. GE : AG :: EH : AB :: FE : FB :: 1 : 3;
.-. AG = 3GE;
.-. AE = AG+GE=iGE;
.: GE^l.AE.
Hence, join the vertex and the centre of gravity of the base,
and the ceiltre of gravity of the solid will be at the distance of
one-fourth of this line from the base.
151. It may be shewn, by a method very similar to the one
in Art. 147, that if four equal bodies be placed in the four
angular points of the pyramid their common centre of gravity
will coincide with the centre of gravity of the pyramid ; and that
the distance of the centre of gravity of any triangular pyramid,
from any plane, is equal to the mean of the distances of its angu-
lar points from the same plane.
152. The line joining the centre of gravity of thfe base
BCD, and that of any parallel section bed of the pyramid being
produced, passes through the vertex A.
153. If a plane be drawn through the centre of gravity of
the pyramid parallel to the base, a fourth part of any line drawn
from the vertex to a point in the base will be intercepted between
this plane and the base.
For a fourth part of AE is intercepted, and therefore "(Eucl.
XI. 16) every line from the vertex to the base is divided in the
same proportion.
154. Hence, if a perpendicular be drawn from A upon the
base, a fourth part of it will be intercepted between the base and
a plane parallel to it through the centre of gravity of the pyra-
mid. And, conversely, if we take a point in the perpendicular
at the distance of one-fourth of its length from the base, a plane
being drawn through that point parallel to the base will pass
through the centre of gravity of the pyramid ; consequently, all
CENTEE OF GRAVITY. 81
other triangular pyramids between the same parallel planes will
have their centres of gravity situated in that plane.
155. To find the centre of gravity of any pyramid.
Let g (fig. 29) be the centre of gravity of the base of the
pyramid ; join Ag. Then, by a method exactly similar to the
one pursued in Art. 150, it may be shjewn that the centres of
gravity of all the plane laminas, parallel to the base, of which
the pyramid may be supposed to be made up, are in Ag, and
consequently the centre of gravity of the pyramid is in Ag.
But we can divide the base BGDEF into triangles, and
suppose the pyramid made up of triangular pyramids, consti-
tuted upon these triangles as bases, and having the common
vertex A. These, by the last article, will have their centres of
gravity in a plane parallel to the base BGDEF, which divides
Ag in G, so that Gg = ^Ag ; consequently the ceritre of gravity
of the whole pyramid will be in that plane, and as it is also in
Ag it must be at G.
156. There is nothing in this demonstration to limit the
number of sides of the base of the pyramid, and therefore in a
cone, upon a curvilinear base of any form whatever, which we may
suppose a polygon of an infinite number of sides, the centre of
gravity will be found, by joining the vertex and the centre of
gravity of the base, and taking a point in that line at the distance
of one-fourth of its length fi-om the base.
157. To find the centre of gravity of the frustwn of a cone
or pyramid cut offhy a plane parallel to the base.
Let BCI) (fig. 30), hcd be the two ends of the frustum,
which are, of course, similar figures; g, g' their centres of
gravity ; G the centre of gravity of the frustum, which will be
in the line gg', because the centre of gravity of every lamina
parallel to the base is in that line. Now complete the frustum
into a pyramid, its vertex A will be in gg' produced (Art. 152) ;
and put a, b for the lengths of coiTesponding parts of the two
ends of the frustum, and c for gg'.
E. s. 11
82 CENTEE OF GRAVITY.
Then Ag' and Ag being like dimensions of the upper pyramid
and the whole pyramid, as are also h and a ; and, because the
like dimensions of similar figures are proportional ;
/, a : 5 :: Ag : Ag' \
.: a : a — h :: Ag : Ag— Ag' = gg' = c;
. ac
a — h'
Similarly, Ag' = rr.
Now, measuring along gA, the distance of the centre of
gravity of the whole pyramid from g = \. r ; and the distance
Ic
of the centre of gravity of the upper pyramid from g' = i. _ , ,
la
and therefore, measuring from g, it =c +i . j-; also, putting
X for the distance of the centre of gravity of the frustum from g,
measuring along gA, we have, by Art. 139,
(whole pyramid) , \ . _ ,
= (frustum) . X + (upper pyramid) . ( c + J . r ] .
But similar solid figures are as the cubes of their like dimen-
sions, wherefore the whole pyramid, the upper pyramid, and the
frustum, which is the difierence between them, are proportional
to a°, V, and a' — V respectively ; and substituting these in the
last equation for the quantities to which they are proportional,
we have
-„.,j,-H-.(.._i,,+j..(,+i._^).
= |.(a'+a=J + a&»+J»-4J»)
GENERAL PEOPEETIES OP THE CENTRE OP GRAVITY, 83
therefore, by dividing the equation hj a — h,
GENERAL PROPERTIES OP THE CENTRE OF GRAVITY.
158. If the mass of each particle cfa system he multiplied hy
the square of its distance from a given point, the sum of the pro-
ducts will he the least possible when the given point is the centre of
gravity of the system.
Let G the centre of gravity of the system be taken for the
origin of co-ordinates ; and put a, h, c for the co-ordinates of the
given point 0; x^ y^s^, x^y^z^...iov those of the particles wjj,
wij... of which the system consists.
Then. ,
( OmX = K - ay+{y, - hy + («. - cf
= ^1 + I/i +. ^1 + a° + &' + c' - 2aa3, - 2hy^ - 2cz^
= ( <ym,)= + {GOy- 2ax^ - 9.hy^ - 2cz^ ;
because (Om^y = x,^ + y,' + z,\ and G(P = d' + h' + c\
"Hence m^{Om^y
= m^.{Gm^'' + m^ . {G0y-2a.m^a. - 2h.m,y^ - 2c.m^z^.
Similarly, m^.{Om^^
= m,.{Gm^' + m,.{G0y-2a.m^,-2h.m^,-2c.m'y„
m^.{Om^^
= m,.{Gm^'' + m,.{GOy-2a.m,x,-2h.m^,-2c.m,z^,
84 GENERAL PROPERTIES OF THE CEHTEE OF GRAVITY.
and consequently, by adding the cotxespolidirig sides of the equa-
tions together,
m, . {Om;f + m, . (Ojw,)' + m, . (Om,)' +
= m,. (<%».)= + mj. (Gfm^r+JW,. (GW,)''+
+ K + »M, + m3+ )-{G-Of
— 2a:{m^x^-^m^^ + m^x^ + )
-ib.(m,y^ + m^^-\rm^^ + )
- 2c . (m,gi +««jSj + OT3S3 + . ...,.•)•
But, because the centre of gravity of the system is in the
origin of co-ordinates, we have, by Art. 1S7,
= mjOJj + wijCCj + OTtjSj +
0=m,yj + 971,^2 + ^3^3+
= »W,^l + Mt/2 + »«A +
Coilsequentiy,
m,. (Omj)' + »w,. (Om/ + m3. {Om^''+
= m,.{Gm,Y+m,.{Om^''-^m^.{Gm^'' +
+ («tj+««>,+m3+ ).{Goy
or, t [m ( Omf] = t{7H{GmY]+%ni.{ (^Of.
From this equation it appears, that the sum of the products
of each particle into the Square of its distance from the point 0,
is greater than X{in{Ghny} by the quantity "tm . GO'; and
since S [m ( GmY} does not depend at all upon the position of
the point 0, the sum will be the least possible when G0 = 0,
that is, when the point is in the centre of gravity of the
system.
159. Cor. 1. So long as the distance of from G remains
the same the quantity X {m {GrmY} +'tm . GC retains the same
value; if, therefore, be fixed in space, and the body be made
to turn round its centre of gravity, the sum of the products of
GENERAL PKOPEETIES OP THE CENTKE OF GEATITr. 85
each particle of the system into the square of its distance from
remains unaltered.
160. Cob. 2. The last two articles are equally true if
7«j, OTj, m, ... he large bodies instead of single particles, observ-
ing, in that case, that x^ y^ a, ,x^y^e^,x^y^z^... wiU be the co-
ordinates of their respective centres of gravity.
161. CoE. 3. Suppose the bodies each equal to m, and let
tlierr number be n, then
S{TO(Omn=OT,.(Om,)= + m,.(OOT,)' + »»,.(Om,)'' +
= m {( OmJ* + ( Ow/ + ( Om.)= + ......}
= m.t{OmY;
and, similarly, %{m{ Gmf] = m . S ( Omf ; also S»i = m^ + m^
+ m^ + =m + m + m-\- to«terms = «m; consequently,
by substituting in the equation of Art. 158, we obtain
m . t {OrrCf = m.%. {Grrif + nm . {G-Of;
. .-. t {OnCf = t . {Qm") +« . {G0)\
It appears then, that in a system of to. equal bodies, the svm of
the squares of the distances of their centres of gravity from a given
point, is greater than the sum of the squares of the corresponding
distances from the centre cf gravity of the system, by n times the
square of the distance of this latter from, the given point.
162. CoE. 4. Hence, if ABG be a triangle, G its centre of
gravity, and a point situated either in the plane of the triangle
or not, we have
A(y + 30" ^ CO" = AG^ ^BGI^^ CG-" -vz . G0\
And in a triangular pyramid whose angular points are A, B,
C, D, and centre of gravity G,
A(J-vBQ-^G(J^B(y
= AG'+BG'+0G^ + DG' + i.GO'.
For, by Art. 147, the centre of gravity of the triangle coin-
cides with that of three equal bodies placed at its angular points ;
86 GENERAL PROPERTIES OF THE CENTRE OP GRAVITY.
and the centre of gravity of the pyramid with that of four equal
bodies at its angular points, (Art. 151).
163. If each particle of a system he ^multipUed, as in
Art. 158, hy the square of its distance from a given point, the
sum of the proiMds will he greater than it would he if the whole
system were collected at its centre of gravity, hy a qvxintity which
is found hy multiplying the products of the hodies taken two and
two respectively, hy the squares of their mutual distances, and
dividing the sum of these products hy the sum of all the hodies.
For let be the given point, G the centre of gravity of the
system of particles or bodies m^, m^, m^... Take for the
origin, and let x, y, z be the co-ordinates of O; x^y^e^, x^y^z^,
x^y^z^,... those of m„ m^, »»,...; also, let [m^m^, {m^m^,
(tojOTj),... be used to denote the distances between j»i and m^,
m^ and m^, m^ and m^
Then, by Art. 135,
5 . 2»ra = »»ia;i + mjjOjjj + jMja;, + ...
'y-tm = m^y^-^ m^^-\-m^^+ ...
s .%m = m^z^ + m^^ + m^^ + ...
squaring each of these equations and adding the results we
obtain
OG\{lmy.^m^.{Om^' + m^. {Om^' + m^' . {Om^' + ...
■ +2m,m^ . {x,x^ + y^y^-¥z^z^ +
+ 2OT,»re, . {x^x^ + y^^ + z^z^ +
+ 2m,w,. Ka33+y^, + a^,) +
+
by writing 0(P, (Om^\ (Om,)', {Om^\.,. for their equals
S'+Z + P, x,'+y,' + z,\ x,' + y^' + z,\ x.^ + y^' + z^ re-
spectively.
But (mjjw,) being the distance between two points whose
co-ordinates are x^y^ z^ , x^y^ s, , we have
GENERAL PROPERTIES OF THE CENTRE OF GRAVITY. 87
{ni,m;)' = K - x,y + {y^ - yj» + {z, - z,y
= ^1 + yi + a.' + a;/ + y/ + «/ - 2 {x,x, + y^^ + z^z^
= ( OmJ' + ( Om^'' - 2 {x^x^ + y,y, + v,) 5
= m^m, {( Omi)» + ( 0>w J - K»»/}.
Smilarly,
2}«j»M, (a;,a;, + 2^j, + 3iS3)
= m^m, {( 0»»/ + ( Om^' - Km,)"].
Consequently, by substitution,
OG' (tmf=^m,\{Om,y + m^. {Om^^ + m^ . {Om,y + ...
+ 7«,7«, {( OmJ" + ( Om,)= - Km/}
+ »Wi'»s {( Om;)'' + ( Omj)^' - Km,)^}
+ m,m3 {( OmJH {Om^Y - {m,m,y}
+
= (»»i+»ij + OT3+ )mj (Ooti)''
+ (mi + »nj + m3+-. )m^{Om^Y
+ {m^+m^+m^ + )m^{Om^Y+
- m^m^ . {m^m^Y- m^m^ . {m^m^''- m^m^ . {m^m^Y- ...
- %n. . S {«i ( Om^] — S \m^^ . {;m^^^\ ;
the term % {»»i»»2 • K"?!)"} feeing understood to represent the
sum of the products of the particles, taken two and two, into the
squares of their mutual distances.
Hence dividing by %m and transposing,
S{m(Omr} = (Sm). 0(?» + ^i»;|?^^,
which expresses the property to be proved.
88 GENERAL PEOPEETIES OP THE CENTRE OP GRAVITY.
164. Cor, 1. From Art, 158, we hare
t {m {OmY} = t{m [GhaY] + tm . {QOf;
which, substituted in the equation above obtained, gives
Am,
A result which might have been obtained at once without
the aid of Art. 158 by supposing to coincide with G.
165. Cor. 2. If now, as in Art. 161, we suppose all the
bodies equal and n in number, the last equation becomes
.•. % ('Wi'mJ =71.% { Qmf.
Hence, in any system i}f n equal bodies, the sum of the squair^
of the lines joining their centres of gravity, two and two, is equal
to n times the sum of the squares of the distances of those points
from the centre of gravity of the system,.
166. Cor, 3. Consequently, in the case of the triangle
(Art. 147),
BG^ ^ AG^ ^AR = Z .{AG^ ^-BG^ + CGf).
Hence the sum <f iJie squares of the three sides of a triangle is
equal to three tim£s the sum of the squares of the distances of its
angular points from its centre of gravity.
167. Cor. 4. In the case of the triangular pyramid we
have
AB" + AC' + A1)'' + BG" + Blf' + GB"
= iiAG' + B(?+GG^ + DCr').
Hence the sum of the squares of the six edges of a triangular
pyramid is equal to four times the sum of the squares of the
distances of its angular points from its centre of gravity.
168. When a system of bodies is in equilibrium under the
action of gravity only, the altitude of the centre of gravity of the
system is in general a maximum or a Tninimum.
General peopekties of the centre of gravity. 89
Let wij, wi„ »i3...be the particles of the system in equi-
librium: z^, »j, «,i.. their respective ialtitudes above a fixed hori-
zontal plane ; z the altitude of the centre of gravity of the system
above the same plane ; g the accelerating force of gravity ; then
"^i^j »»j5'> »»s5'-" ^^^ tlie forces acting upon the particles of the
system. Let now the system be disturbed in a manner subject
to the same restrictions as were pointed out in the Chapter on
virtual velocities, (i. e. rods must not be bent, cords must be
kept of invariable length, contacts must not be broken, «S;c.) and
let (?a,, dz^, <fo3...be the virtual velocities of the respective par-
ticles, then by Art. 113, because there is equilibrium,
m^g . dz^ + m^g.dz^ + m^g.dz^+...=0,
or % (mdz) = 0.
But since "Zm .1 = X (rnz) ;
.'. Xm . dz = S (mdz) = 0, .". dz = 0.
Now dz is the differential of i, or second term of Taylor's
Theorem, and this being equal of zero, it follows that i is in
general a maximum or minimum.
It has been stated that the principle of virtual velocities ex-
tends only of quantities to the first order of smallness, that is, to
the second term of Taylor's Theorem only ; the equilibrium
of the system therefore does not require that S {md'z) shall be
equal to zero, though it may happen to be so in particular cases ;
and the algebraic sign of d^ will decide whether s is a maximum
or a minimum.
169. Cor. Since the centre of gravity of the system is the
point through which the resultant % {nig), or g'Zm of all the
forces m^, mj,^... passes, and seeing that this resultant acts in a
downward direction, it appears that, if the system be disturbed,
the tendency of gravity is to make the centre of gravity descend :
but if the geometrical constitution of the system be such that in
passing out of a position of equilibrium the centre of gravity can
only ascend, the ascent will be opposed by gravity ; that is, in
this case gravity tends to bring the system back again into its
position of equilibrium. But if the constitution of the system be
E. s. 12
90 GENEEAL PROPERTIES OF THE CENTRE OP GRAVITY;
such tli9,t in passing out of equilibrium'the centre of gravity can-
not tut descend, it is assisted in its descent ty gravity, ajid there
is na tendency to return totwai?ds the position irom which it set
out, HeiBQe it follows :
(1) That if the altitu,de of the centre of gravity be a
minimum., the system when disturbed will return by the action of
gravity towards the position from which it was disturbed. This
is therefore called a position ot stable equilibrium.
(2) That if the altitude of the centre of gravity be a maxi-
mum, the system when disturbed will recede by the action of
gravity still farther from the position of equilibrium. This is
therefore called a position of unstable equilibrium.
(3) That if the centre of gravity neither ascend nor descend
when the system is disturbed, it still continues in a position
of equilibrium. This is therefore called a position of neuter equi-
librium.
170. If a body be placed with its base upon a plane it will
stand or fall according as a vertical through its centre of gravity
falls within or without its base.
Let AB (figs, 31, 32) be the base of the body, G its centre of
gravity ; draw a vertical through G meeting the plane on which
the body is placed in H; H falling within the base in fig. 31,
and without it in fig 32.
Every particle of the body is acted on by the force of gravity,
and we have shewn that the centre of gravity is the point at
which the resultant 'of the forces may be supposed to act : this
resultant is equal to their sum, that is, it is equal to (TF')the
whole weight of the body. We may therefore suppose the body
to be without weight, and that a force acts at G equal to W. la
.fig. 31, we may suppose this force to be transmitted to H, which
being in contact with a fixed point of the plane cannot be moved»
and therefore W is counteracted, its effects being to mak& the
body stand firm upon its base. But in fig. 32, TT cannot be
transmitted to a point which is in contact with the plane, and
therefore as there is nothing to oppose its action, 'the point O
GENERAL PROPERTIES OF THE CENTRE OP GRAVITY. 91
will descend, thereby causing the body to turn about the
point A.
171. This reasoning applies if the plane on which the body
is placed be not horizontal, provided the body be prevented from
sliding by the roughness of the plane, or any equivalent cause.
172. If a body be placed on points, instead of a flat base, it
will stand or fall according as a vertical through its centre of
gravity falls within or without the polygon formed by passing a
thread round the points.
173. If there be any case not here considered, it may be
disposed of on the following principle. The whole weight of the
body may be supposed to act at its centre of gravity ; and as it
acts in a downward direction, its tendency is to cause that point
to descend. If the geometrical arrangement of the system be
such that it is impossible for it to move so as to permit the centre
of gravity to descend, it will remain stationary ; for in this case
the tendency which gravity produces is prevented from taking
effect from the construction of the machine.
APPLICATION OF THE INTEGRAL CALCULUS TO FIND THE
CENTRE OF GRAVITY OF BODIES.
174. To find the centre of gravity of a plane curve tine.
Let AB (fig. 33) be the curve line, referred to the rectangular
axes Ox, Oy. P any point in AS, and Q very near to P.
x= OM, Bx==MN, y-MP, s = AP, Bs = PQ, u=ithe moment
of the arc AP, and Sm = that of PQ, about Oy.
The moment of PQ about Oy is greater than it would be if
PQ were all collected in a point at P;
/. &u>x8s;
and it is less than if PQ were all collected at Q ;
.■. Su< {x + Bx) Bs.
92 GENERAL PROPERTIES OF THE CENTRE OP GRAVITY.
Hence j- always lies between x and x + Bx, consequently
the limit of -«- = a; ;
but by the principles of the Differential Calculus
the limit of -=r- = t" 5
OS as
du _
'■•■^"'"'
/, u — Jxds,
the integral to be taken from x=OC to x= OD.
But if xy be the co-ordinates of the centre of gravity of AP, we
have by Art. 1S9,
xs = u=jxds',
- Jxds
Similarly, y=^^—.
175. These formulae will suffice for the determination of the
point required in any given example: but it may be remarked
with respect to these, and other formulsa, which will be investi-
gated for finding the centres of gravity of areas and volumes,
that they are not always of convenient application. It is, gene-
rally speaking, more easy to work out an example by taking an
element 8m of the figure, and then applying the equations
_ _ X {xZm) -_'%{yhn)
If this method be applied to the case investigated in the
last Article, we have 8»» = 8s; .*. ^ (Sm)=SSs=/tfe = s; and
% {xSm) = S {xSa) =fx8sj ; and .'. x = — , the same result as
before
GENEEAI. PEOPEKTIES OF THE CENTRE OP GEAVITT. 93
Ex. 1. To find the centre of gravity of the arc of a semi-
cycUnd.
Let BG (fig. 34) be the base, AB the axis, and ^C-the arc
of the semi-cycloid; x = AM, y = MP, 8 = AP, 2a = AB; then
the equation of the cycloid is
If = {2ax — ar')' + a vers"* - ;
and ds= ( — J dx;
.'. s = 2 V2aa;.
Also, xds = '^2ax.dx',
.'. Jxds = -x^f2ax^,
o
.*• 5=1 .(1).
Again, fyd8=ys-Jsdy
= ^5 - J]2 V2aS f — - 1 j cfe
= ys - 2 V2a/(2a - a;)* dx
= 2^« + |V2a(2a-a!)'+a
Now this integral ought to vanish when x = 0;
••• C^=-T"'
and fyd8=ys + -^ '/Za {2a - x)' - — a' ;
. 5:^„ ,H(2lZ^ 8^ (2).
••^^^3 V5 3V2^
94 GENERAL PROPERTIES OF THE CENTRE OP GRAVITY.
Tte equations (1), (2) give tlie co-ordinates o£ the centre of
gravity of any arc AP: and if we write in them 2a for x, we find
|aiid(^-|)a,
for the co-ordinates of the centre of gravity of tlce arc A G.
Ex, 2. To find the centre of gravity of an arc of a circle.
Let AB (fig. 35) be the given arc, its centre, G its middle
point; join OA, OB, OG: and letP^ be a very small element
of the arc. Draw Oy perpendicular to OG. a= OA, a =-4 0(7,
6 = GOP, h6 = POQ : the centre of gravity oiAB is manifestly
in the line OG, let x be its distance fi'om measured along OG.
Then
the element PQ = aSd,
its moment about Oy = aSd .acos0;
.: moment of the arc ^5= a" /cos Odd feom 0=: — ato0 = + a
= a" sin 6, irom 6 = — at<y0 = + a,
= 2a'' sin a,
and arc AB = 2aa. ;
- _ S {a^m)
_ 2a^sina
2aa.
sin a
= a ■
Ex. 3. The equation of a catenary being
V 2 2
l^He' + e"),
and a;y being the co-ordinates of the semi-arc (s), shew that
- oi ay _ aon
GENERAL PROPERTIES OP THE CENTRE OP GRAVITY. 95
Ex. 4. The equation of a parabola being y^^Amx, shew
that the distance of the centre of gravity of the arc, cut off by the
latus rectum, from its vertex, is
m 3 V2- log, (1+^/2)
^* V2 + log,(H-V2) ■
176. To find the centre of gravity of a plane area.
Let A CDB (fig. 33) be the area : then using the same nota-
tion as in Art. 174,
an element of area = PN= yZx,
its moment about Oy = x. ydx ultimately ;
.*. moment of the whole area about Oy = % {xyBx) ;
=j0ydx,
and the whole area = fydx ;
- _fxychii
~ fydx ■
The integrals are to be taken from x= OG to x= OB.
Again,
the moment of the element ySx about Ox
= ydx . I ultimately ;
.•. moment of the whole about Ox = ^fy'dx;
'•y-^Sydx:
Ex. 1. To find the centre of gravity of the area of a serm-
parabola.
Let A (fig. 34) be the vertex, and AB the axis of the
parabola ; and let y^ = 4mx be the equation of the curve, where
x = AM, a,ndLy = MP; iput AB = a;
.'. Jxydx = J'J4,m,x'dx
= - V 4m .x^+
5
o
= - 'Jim . a^ between x = and x = a.
o
96 GENERAL PEOPEETIES OF THE CENTRE OP GRAVITT.
Also ^i/^dx = jimxdx
— 2mo? between the same limits :
and jydx = j ^^inx^dx
= -r sT^ka* between the same limits ;
- fVima^ 3 3 -„
.•. ai = - ° ■ . =-a = -=AB;
I V4»wa* 5 5
, _ 1 awia" 3 J 3 „ „
Ex. 2. To find the centre of gravity of the area of a circular
sector.
Let A OB (fig. 35) he the given area, x the distance of its
centre of gravity from ; then using the same notation as in
Ex. 2 of Art. 175, we have
elementary area = APOQ — ^a'SO ultimately ;
.•. area of the sector =j^c^dd from 6=— a to = + a
Now the elementary area FOQ being ultimately a triangle,
we may suppose its centre of gravity to be at g, such that
Off = % OP=%a: and as the distance of g from Oy — %acos0
ultimately, we have the moment of the elementary area QOP
about Oy
= - c^W .-a COS 6;
.'. moment of the sector about Oy
= 1 a' J cos Odd
3 •'
= -a°sin5+ C
o
2
= ^ a' sin a from 5 = — ato^ = + a;
CENTRE OP GEAVITY OF A SOLID OF REVOLUTION, 97
fa' sin a
.T=-
'2a sin a
3a '
Ex. 3. If xy be the co-ordinates of the centre of gravity of
the area of a semi-cycloid whose equation is
y={2ax-xy + avera-'-,
- 7a , _ air r 1 6 \
Ex. 4. If xy be the co-ordinates of the centre of gravity of
the area cut off from a parabola {if = 4ma3) by a focal chord in-
clined to the axis at an angle a,
« = — (3 + 8 cot^ a) and y = 2m cot a.
Ex. 5, To find the centre of gravity of the area of the
quadrant of a circle, whose equation is ic^+y = a"
- 4a _
Ex. 6. To find the centre of gravity of th^ node of the
lemniscate, whose equation is r' = a" cos 20,
177. To find the centre of gravity of a solid of revolution.
Let AB (fig. 33) be the curve by the revolution of which
round Ox the given solid is generated. Make the same construc-
tion and notation as before. Let V denote the volume of the
solid generated by the revolution of AMP, and S V that generated
by PMNQ ; m = the moment of F round Oy, and hi that of SF
about Oy.
The moment of SF about Oy is greater than it would be
E. s. 13
98 CENTRE OF GRAVITY OF A SOLID OF REVOLUTION.
if SF were all collected in the circular plane generated by PM,
that is,
hu>x.hV;
and it is less than it would be if SFwere all collected in the
circular plane generated by QN, that is,
Sm < (a; + Sa;) . S V.
Hence ,c- always lies between x . k— and x.^- + S F.
Whence, as in Art. 174,
dx ' dx'
.-. u^J{xdV).
But X, y being the co-ordinates of the centre of gravity
of F,
x.V=u = S{xdV).
Now dV= nry^dx, by the Differential Calculus ; and, therefore,
V='7rji/'dx; consequently
X Ji^dx = J xy'dx ;
- _Jxi/^dx
" /y<^x •
From A'rt. 143, it is manifest that
y = o.
Ex. 1. To find the centre of gravity of a hemisphere.
A hemisphere is generated by the revolution of a quadrant
whose equation is
y = 2aa;' — x ;
•■• Sy'dx = ax'--x^,
which gives, for the whole hemisphere, by writing a for x, the
2
3*
2
quantity -a'.
CENTRE OF GRAVITY OP A SOLID 0¥ ANY FORM. 99
Again, Jxy'dx = / {2aas' - x') dx,
2 3 1 1
which, by writing a for x, becomes — a* ;
1^
x =
5 5
= o a = - of the radius,
o o
Ex. 2. Given the altitude (c) and the radii (a, 5) of the ends
of a parabolic frustum, to find its centre of gravity ;
* = 3-^^T6^' ^^^2^ = =
a; being measm-ed along the axis from the smaller end whose
radius is a.
Ex. 3. In a cone, generated by the revolution of a right-
angled triangle about one of its sides,
5 = f of that side.
Ex. 4. In the solid formed by the revolution of a semi-
cycloid about its axis,
__« 637r^ - 64
'*'~6" 97^-16 ■
X being measured from the base along the axis.
Ex. 5. In the paraboloid, formed by the revolution of the
parabola, whose equation is y"''*" = a^a;".
_ m + Sn X
m + 2n' 2 '
178. To find, the centre of gravity of a solid of any form.
Let Ox, Oy, Os (fig. 36) be the rectangular co-ordinate axes
to which the solid is referred by its equation. Let ASPG be a
portion of the surface of the solid, comprehended between the
100 CENTRE OP GEAVITY OP A SOLID OP ANY POEM.
co-ordinate planes xOz, yOz, and the planes PpNG, PpMB re-
spectively parallel to them. Through the point S very near to
P draw planes 8snc, Ssmh parallel to the former. Let xyz be the
co-ordinates of P, and x-'r^x, y + By, z + Se those of 8. Then,
denoting the volume of the parallelepiped Ps hy A, its moment
about the axis Ox is greater than if it were all collected in the
plane Pq, and less than if collected in the plane Us ; that is, the
moment of A is
greater than yA,
and less than {y + Sy) A.
But now if u be the moment of the solid PO about Ox, the
moment of BBmPn about Ox will be (by Taylor's theorem ap-
plied to two variables x, y)
djui . Bx -\-\d^u . {SxY + ...
dyU . hy + d^dyU . SxSy + ...
+ ^d,'u.{Syy+...
+ ...
and by the same theorem, applied to the variable x, the moment
of the solid BmP. ahout Ox is
d^u . Sx + ^d^u . {SxY + ...
and, similarly, the moment of the solid CnP, is
d^u . Sy + \dyU . {SyY + ...
Subtracting both these from the former, we find the moment
of the parallelopiped Ps to be equal to d^d^u.SxSy+ ...; conse-
quently, this quantity always lies between yA and {y + Sy) A ;
and, therefore, dji^u + ... always lies between
■^°^ SxSy ^®^^® ^° ^ ^^ ^*^ ^™^*' ^^^ consequently the two
quantities 2,. g^ and 3/. g|-+S3,.^tend to equality with
ys; and d^d^u+ ... which always lies between them, tends to
CENTRE OF GRAVITY OP AN ELLIPSOID. 101
d^d^u as its limit ; the three limits are therefore equal ; conse-
quently,
d^u^yz;
•■• u = Uy[yz).
Now the volume of PO is equal to jj^z, and its moment
about Ox is
wherefore, by Art. 139,
y-U^=Uv{y.^) (2).
By a similar investigation, we should find
«-/J»«=/J»M- (!)•
And observing that the centre of gravity of the parallelepiped
A is ultimately in its middle point, we should find
^•U« = 4U(^') (3).
Remark. It is evident, that by taking an elementary
parallelepiped, at right angles to the plane xOz, we might
also obtain
^•/J«2/=/J«(«y);
and if the elementary parallelopiped were at right angles to the
plane y Oe, we should find
y-!y!>«i = !yh{«>y)>
« •/!,/««= /J«M-
These formulae are in fact, often more convenient than those
first given; and which are the most convenient in a given
example is to be determined by the form of the body and its
situation with respect to the co-ordinate planes ; the choice must,
however, be left to the skill of the reader, as no general rule can
be laid down. In every case, the greatest care is requisite to
take the integrals between proper limits.
102 CENTRE OP GRAVITY OF AN ELLIPSOID.
All the three sets of formulje are comprehended in the
following : —
which may be readily investigated after the manner of Art. 175.
Ex. 1. To find the centre of gravity of the eighth part of an
The equation of the surface of the ellipsoid is
_2 "r 12 "r _a '■•
+ (7
This integral is to be taken from y = 0, to that value of y
ich mi
therefore
which makes » = ; or from y = 0, to y = - Va* — a;'* ; and
CENTRE OF GRAVITY OF A SURFACE OP REVOLUTION. 103
This integral is to be taken from x = 0, to a; = a ; and
therefore
Again, to find the value of J^Jy {xz) we observe that
ly{xz)=xf^z
bcir . i 2-
••• /J. (fl'«)=:5 •/.(«'«; -a;')
which, taken between the same limits as before, viz. x = 0, and
x = a, gives
iTifbc
hL (aJ^) = ■
Hence x
16 ■
16 '
_ 3
a; = -o.
— 3
Similarly, y = -^b;
and z =xc.
o
Ex. 2, To find the centre of gravity of a portion of a
paraboloid, comprehended between two planes passing through
its axis at right angles to each other.
If a be its length, and h the radius of its base, the co-ordi-
nates of its centre of gravity will be
_ 2 _ _ 16&
104 CENTRE OF GRAVITY OF A SUEFACEOP KEVOLUTION.
179. To find the centre of gravity of a surface of revo-
Employing the notation and figure of Art. 177, let u be the
moment of the surface generated by the arc AP, and therefore
Bu the moment of- that generated by PQ; let 8 denote the
former, and 88 the latter of these surfaces so generated. Then
the moment of 88 about Oy is greater than if it were all col-
lected in the circumference of the circle described by P, and less
than if collected in the circumference of that described by Q,
that is,
Bu is greater than x . B8, and less than (x + Bx) . B8;
.•. c^ lies between x ^- and x ■^- + BS.
bx ox ox.
Equating the limits, as before, we have
du
dx
d8_
dx
.
•. u = 2Tr J{xyds).
But
u =
the moment of 8 about Oy
= x8= X . 2irf{yds) ;
.'. X.
^TrJiyds)
= 27r J{xyds) ;
,',
■xfiyds) =
= J{xyds).
And it is evident, from the symmetrical form of the surface,
that ^ = 0.
Ex. 1. To find the centre of gravity of the surface of a
cone.
If a be the altitude and h the radius of the base of the cone,
the equation of the line by which the surface is generated is
y=-^
hx
' = (i + -s) <^ ;
CENTRE OP GRAVITY OF A SURFACE OF REVOX,UTION. 105
■which, taken between the limits x=0, and x = a, gives
Also,
3a V
bx' /, by „
which, between the same limits, gives
J{xyds)=iab^^+¥;
.: x.^b'/J+¥ = iab'/^ni'',
Ex. 2, To find the centre of gravity of the surface generated
by the revolution of an arc of a circle about a diameter.
The centre of gravity bisects the axis of the zone.
Ex. 3. To find the centre of gravity of the surface generated
by the revolution of a semi-cycloid about its axis,
Ex. 4. To find the centre of gravity of the surface of
a paraboloid.
Taking the focus as origin of polar co-ordinates, we find the
distance of the centre of gravity from the directrix
Q sec' - - 1
3m 2
sec'--!
E. S.
14
106 CENTKE OF GRAVITY OF A SUEPACE OP ANT POEM.
Ex. 5. To find tlie centre of gravity of the surface generated
by the revolution of a node of the Lemniscate about its axis.
a 1 - cos' ie a 2 V2 - 1
aj =
6 ■ 1 - cos 12 ■ V2 - 1
180. To find the centre of gravity of a surface of any
form.
If, in Art. 178, we use A to denote the elementary surface P8
instead of the prism Ps, we shall have
the limit of ~- = \/TT{d;^f+Wf ;
and by proceeding exactly as in that Article, we shall find
X . u ^^1 + {d^y + id^f=Uy {« ^^1 + {d^r + iA^)\
z.Uy >/T+Wf+Wf=Uy {« '^i + {dj,r+{d,zy}.
181. To find the centre of gravity of a curve of dovMe
curvature.
If we use 8 for the length of the curve line, and hS for
the length of a very small portion of it, we shall have the
so
limit of s- = d^8 = Vl + [d^yf + {d^Y, and it will be found
that
x8=jjx>^/\ + {d^yy+{d,zy,
y8=J.y'Jl+Jd~yy+Wr,
z8 = !^^/l + {d^yy+{d^zy.
182. We shall now add a few examples of finding the centre
of gravity when the density is variable. Questions of this kind
depend upon the formulae of Art. 138, viz. —
53 tjpVx) , - tjpVy) , -_{tpVzl
''-%{pV)^ ^-t{pV)' '-t{pV)-
CENTRE OP GEAVITY OP A LINE. 107
183. To find the centre of gravity of a physical line, the
density of which at any point varies as the n* power of its
distance from a given point in the line produced.
Let AB be the given line, and C j j-y-
the given point ; n = the density at a '^ •* ^
point in AB, whose distance from C=l; a= CA, b= CB,
X = CP, hx = PQ. Since a physical line is of uniform thickness
throughout, we may take the length of any portion of it as the
measure of the volume of that portion ; hence hx = the volume
oi PQ, and as the density varies as (distance from 0)";
.'. r : a;" :: /* : fix".
Wherefore the density at P is /ao;", and PQ is ultimately
of uniform density, therefore the mass of PQ is
= fix^hx ;
.-. the mass of AB= S {fuxfSx)
= fi% (af'Bx)
= fi,Jx''dx
= li
= '*• n + l '
between the limits x = a and x = b.
Again, the moment of the mass of PQ about C
= lj,x"-'^Sx;
.: moment of AB about C = fijaf*^dx
x"^'
+ G
n+l
&"«-
«««
= fi.
^ +G
n + 2
"''*■ w + 2
between the same limits as before.
108 A TRIANGLE OF VARIABLE DENSITY.
Wherefore x being the distance of the centre of gravity of
the line from 0, we have
-_ t{pVx)
_ n+l h'^-a"'^
Eemakk. When m = — 1,
M/X = /*/j
= fi . logjo; + G
And ,ijjc'"-^ = ,j.{b-a);
- h — a
.'. x =
Again, when m = — 2,
and fji,Jx"*'^dx = fj, I
dx
X
1 ^
= /*log.-;
a5 , 6
Ex. 2. To ^^jm:? <Ae centre of gravity of a triangular plate, of
uniform thickness, the density of which at any point varies as
the n*** power of its distance from a line through the vertex
parallel to the hose.
A TRIANGLE OP VARIABLE DENSITY. 109
Let ABC (fig, 37) be the triangle, OD a line through
its vertex parallel to its base; /* the density at a point in
the triangle at the distance 1 from OD ; P, Q two points
in ^C very near each other, through which draw Pp, Qq
parallel to the base ; h = AG, c = AB, x — OP, hx = PQ,
e = ^GAB=iACD.
Then the density at every point in the line Pp = ii{x sin 6)",
which may be ultimately taken as the density at every point of
the element Pq. We may regard Pq as a parallelogram, whose
base I)p
= -Y" , by similar triangles A CB, POp ;
and whose altitude is PQ sin = ^x.B\rx6; its area, which we
may take as the measure of its volume, is therefore
= -T- . 8a; . sin 6 •
and its mass
= fiix sin 6)" .r. xSx . sin 6
= ^(xsin0)"«8a;;
.-. the mass of the triangle
= %if^{xsiner'Bx^
= ^.(sin6T7a;"-'Va;
= ^. (sin ^r.^+C^
h ^ n + 2
■Ln-n
= ^.(sin0r".-
b ■ ^ ' n + 2
n + 2 ^
110 A QUADEANT OP VAEIABLE DENSITY.
And the moment of the element Pq about CD
= ^(a;sin0)"-*^Sa;.
Therefore the moment of the triangle about CD
= ^/(a;sin0)"«Ja;
= ^°(sin^)"«.^^+C
= ^(8in^r.-^
o ^ ' n+ 3
n + 5
Wherefore, if a line passing through the centre of gravity
of the triangle, parallel to the base, cut -4 C at a distance x from
C, the distance of the centre of gravity from CD will be x sin 0,
and
.'. a; sm 6/ =
-^ . (h sin 0)"«
- « + 2
.'. 03 =
n + 3
.AC.
And if CE be drawn bisecting the base, the centre of gravity
must be I in that line ; hence we have two lines passing through
the centre of gravity, and consequently it is the point of their
intersection.
Ex. 3. To find the centre of gravity of a quadrant of a circle,
the density at any point of which varies as the n* power of its
distance from the centre.
A QUADRANT OP VARIABLE DENSITY. Ill
Let ABC (fig. 38) be the quadrant; CD, Cd two radii
making angles with CA respectively equal to 6, 6+h6; AO=a,
CP= Cp = r, PQ = pq=hr; /i = density at the distance 1 from
the centre ; therefore the density at P or ^ = /ir". Now we may
ultimately consider Pq as a parallelogram, whose sides are PQ
and i^, or Sr and rhO, and its area = rBr . Bd, which may he
taken as the measure of its volume ; and its mass
= fir'' . rBr . SO ;
.-. mass of the quadrant = /«/, {fir"*^).
A' „n+2 I rt
Now/,(/.0 =
w + 2
a""^ from r = to r = a.
n + 2
mass of the quadrant = —r^ • /»«'
= -^ . ar'^e+ 0, from 0=0 to 5 = |
n + 2 ' 2
/* T „n+2
n + 2 2
Again, the moment of Pq about OB
= fjur" . rSr .B9.rcoa9;
.-. moment of quad, about OB = J»f^{jJt,r"'^coa0).
But /, (a'»-"'^ cos 6) = -^ . r"« cos ^ + G
■""^ ' n + 3
n + 3
moment of quad, about CB = ^^^ . /««"''' cos ^
= -^.a''*'sin^+C
n + 3
112 SPHERICAL WEDGE OF VARIABLE DENSITY.
between the same limits as before ;
m + 3
03 = -
'* '^.a"
w+2 2
_n+2 2a
n + 3' it'
And it is manifest, from the symmetrical form of the figure,
with regard to GA and GB, that y = x.
Ex. 4. A sector of a circle ACB (fig. 39) revolves round one
of its radii AC thraugJi a given angle (jS), and generates a solid,
the density at any point of which varies as the (n)**" power of its
distant from the centre C ; to find the centre of gravity of the
Since the solid is perfectly symmetrical with regard to a
plane passing through A G, and bisecting the angle /8, the centre
of gravity must be in that plane. Let GA be the axis of x, and
a line in the plane BGA at right angles to AG, the axis of y ;
the axis of a being at right angles to both ;
.-, z=y tan-.
Leta = ^C, a=^BGA, = EGA, Sd = FGE, GP= Gp = r,
PQ =pq = hr, fi = the density at the distance 1 from O. Then
the area of the parallelogram Qp
= rBd . Br ;
and when the sector revolves about A C, this parallelogram gene-
rates a volume
= r sin 6 . ^'. rBd . Br
= 0f^Br.smeBd;
for P's distance from AG is r sin 9, and in revolving through the
angle /3, the length of its path is r sin ^ . y8. The density of
this volume
SPHERICAL WEDGE OF VARIABLE DENSITY, 113
and therefore the mass of the element generated by Qp
= fir'-. ^r'Sr. sine. Be-,
.". the mass of the solid = /^/S /g/^r"'^ sin 9.
But /,,?•"« sin e = -^— .aine+C
n -f- o
-.11+3
. sin d,
m + 3
from r = to r = a ;
.*. the mass of the solid = . a^'^Je sin 6
- '^^ .a-'+'cos^+C
m + 3
t + 3
.a''^'(l-cosa)
"^ .a"*' sin*
« + 3 2'
from = to 5 = a.
Again, the moment of the elementary mass with respect to
the plane yz
= H^r"-^ sin .Sr .B9 .r cos ;
.•. the moment of solid = /J'^Jefr {r"'*'" sin cos 0)
- '*^ a"^7<, (sin 008 0)
w + 4
n + 4:
,o"«sin=0+ C
= i,Jf^.a''«sin'a;
^ w + 4
_ , ji + 3 sin' a
* w + 4 . »«
sm-
w + 3 -a
= . a cos - .
n + 4 2
E. S.
15
114 SPHERICAL WEDGE OP VARIABLE DENSITY.
In order to find s, we must divide the volume generated by
tlie revolution x)f the parallelogram Pg into elements ; to this
end, let there be two planes passing through A C and inclined to
the plane BGA, at the angles </> and ^ + S^ respectively; then
the portion comprehended between them will be equal to the
volume generated by Pq, in revolving through an angle S^, and
therefore is
= r sin 9 .B<}). rB6 , &•
= r'Sr . sin 689 . S^. .
And the density of this element is /lir", and therefore its mass is
fir^'^'Sr. sin 989. S4>,
and its distance from the plane ABO is r-sin^.sin^, as is evi-
dent from the construction J and therefore its moment with
jespect to the plane ay
= /ir"+»S»- . sin" 9S9 . sin <j)Bcf, ;
therefore the moment of the solid with respect to the plane ajy
= /*/J»/« (»•"'" sin' sin ^)
11+4
= ^.Ieh{sin^esini>)
= ^!B{-ain'6cos6+C)
n+4
= — 4/«(l-cos;8)sin'^,
taken from ^ = to ^ = /8.
Now /» sin" = J/« (1 - cos 20)
= \{9-^sin26)+ G
_ a sin 2a
~2 ^'
taken fi-om ^=0 io 9 = ol;
SPHERICAL -WEDGE OP VAEIABLE DENSITY. 115
therefore the moment of the solid with respect to the plane m/
n + 4
sin" -x (a — sin a. cos a) ;
■ ,/8
, „ sm"^
, - _ w + 3 a 2 a — sm a cos a _
■■^~w + 4"2' . ,a is '
— — B
and therefore w = s cot —
^ 2
B . ^
, „ cos - sm ^
_n + S a 2 2 a — sm a cos a
~w + 4"2* . jK ■ is ■
sm"-
Ex. 5. Find the centre of gravity of a cone, the density at
every point of which varies as the square of its distance from a
plane through the vertex parallel to the base.
5 = - of the cone's axis.
6
Ex. 6. Find the centre of gravity of the eighth part of a
sphere, the density at any point, whose distance from the centre
is r, being proportional to
a . irr
-sm— -,
7- 2a'
where a denotes the radius of the sphere.
5 = y = i = a(Jl_y.
116 guldin's peoperties.
guldin's properties.
184. The surface generated hy a plane curve line, which re-
volves about a fixed axis, is equal to the product of the length
of the curve line by the length of the path described by its centre
of gravity.
For let AB (fig. 33) be the curve line, and Ox the line about
which it revolves through an angle 0; then using the same
notation as in Art. 174, the point F describes an arc =yd, con-^
sequently the arc FQ describes a zone, of which the length is
y9 ultimately, and the breadth = Ss ; hence the area of the zone
is ultimately = 6yBs ; and therefore the area of the whole surface
generated is
= -Z{dySs) = 0Jyds;
the integral being taken between the limits corresponding to
x=OC, x=OI).
But if y be the distance of the centre of gravity of the arc
AB from the axis Ox, we have shewn in Art. 174, that
y . (arc AB) = J yds, between the same limits ;
hence the surface generated
= 0y . (arc AB).
Now 0y is the length of the path described by the centre of
gravity, consequently the last equation expresses the property
to be proved.
185. The volume generated by a plane area, revolving ahout
a fixed axis in its own plane, is equal to the product of the area
into the length of the path described by its centre of gravity.
Let A be the revolving area; hA a portion of it so small
that it may be all considered to be at the same distane y from
the axis. Then if ^ be the' angle through which the area re-
volves, BA will describe a volume which may be considered to
be a thin cylinder bent into the form of a portion of a ring.
The area of the base of this cylinder is hA, and its length is y0,
consequently the volume generated by hA
= 0yhA;
guldin's properties. 117
and therefore the whole volume generated
But if y be the distance of the centre of gravity of the
area A from the fixed axis, we have from the nature of the
centre of gravity
X{BA).y = t{yBA),
or Ay = %{yhA);
hence the whole volume generated
= eyA:
an equation which expresses the property which was to be
proved.
Remark. If the curve line in Art. 184, or the plane area in
Art. 185, does not revolve about a fixed axis during its whole
motion but moves in any such manner that it may at any moment
be assumed to be revolving for an instant about a fixed axis in its
plane; then the propositions in those articles will be true for
each instant; and consequently, by adding these results together,
those articles mil he true for the whole motion whatever be the
nature of the path of the centre of gravity. But it is necessary
to notice that when the instantaneous axis, about which the
generating curve or area is supposed to be revolving, is in such
a position that the instantaneous axis divides the curve or area
into two portions, the part generated by one of those portions
during that instant is to be considered positive, and that gene-
rated by the other negative, and the propositions fail in this case.
As long therefore as the line of instantaneous revolution lies en-
tirely out of the limits of the generating curve or area, the pro-
positions in Arts. 184, 185 hold true, viz. :
The surface generated hy a plane curve line which moves in
any manner (subject to the limitations just named), is equal to
the product of the length of the curve line hy the length of the
path described hy its centre of gravity. And
The ■volume generated hy a plane area, which moves in any
wiawwer (subject to the same limitations), is equal to the product
of the area into the length of the path described hy its centre of
gravity.
CHAPTEE VII.
ON MECHANICAL INSTRUMENTS.
186. Every machine, how complicated soever its con-
stmction, is found to be reducible to a set of simple ones,
called the Mechanical Powers. These, though authors differ
considerably on the subject, are generally said to be six in
number, viz.:
1. The Lever;'
2. The Pulley;
3. The Wheel and Axle;
4. The Inclined Plane;
5. The Screw;
6. The Wedge.
These are not the most simple machines ; for, rods used in
pushing, and cords used in pulling, are much more simple ; in
fact, every machine will be found to be a combination of levers,
cords, and inclined planes, and these might consequently be
called the simple Mechanical Powers, with much greater pro-
priety than the six before mentioned. As, however, these are
not very complicated in construction and application, and as
levers, cords, and inclined planes do always, in actual practice,
present themselves in machinery, in one or more of these six
combinations, it will very much facilitate our enquiries into the
properties of any proposed machine, to be acquainted with their
forms and the advantages to be expected from their use.
In speaking of any machine, the force which is applied to
work it is called the Working Power, or simply, the Power; the
weight to be raised, or resistance to be overcome, is called the
Weight; the point where the machine is applied to produce its
effect is called the Worhing Point; and the fraction
Weight
Power
MECHANICAL INSTEDMENTS. 119
is called the Mechanical Advantage (by some authors the Power,
but this creates confusion by confounding it with the former
definition of power) of the machine.
187. Every machine is useless until put in motion, and
therefore its parts ought to be so arranged and adapted that
the given power may be able to overcome the proposed weight,
and move it with the requisite degree of celerity ; but, in dis-
cussing the theory of the Mechanical Powers, it will be suffi-
cient to determine the ratio of the weight to the power when
they balance each other, for then the slightest addition made to
the power will cause it to preponderate and put the machine in
motion.
188. It is veiy important to remark, that when a power is
employed in working a machine, a very considerable portion of
it is found not to reach the working point, being spent in over-
coming the stiffness of the cords and the roughness of surfaces
whichjnib against each other. Much power is also lost through
the imperfection of workmanship, the bending of rods, beams
and other materials, which are intended to be rigid, the resist-
ance of the air, &c. ; but the introduction of the consideration
of these things, though very important in a practical point of
view, would only tend to embarrass the student by rendering
our investigations tedious and perplexing. We shall therefore
at first suppose cords to be perfectly flexible, surfaces quite
smooth, workmanship geometrically exact, rods and beams per-
fectly rigid, the air to offer no resistance ; &c.
" It is scarcely necessary to state, that, all these suppositions
being false, none of the consequences deduced from them can
be true. Nevertheless, as it is the business of Art to bring
machines as near to this state of ideal perfection as possible, the
conclusions which are thus obtained, though false in a strict
sense, yet deviate from the truth in but a small degree. Like
the first outline of a picture, they resemble in their general
features that truth, to which, after many subsequent corrections,
they must finally approximate.
" After a first approximation has been made on the several
suppositions which have been mentioned, various effects, which
120 MECHANICAL INSTRUMENTS,
have been previously neglected, are successively taken into
account. Eoughness, rigidity, imperfect flexibility, the resist-
ance of air and other fluids, the effects of the weight and inertia
of the machine, are severally examined, and their laws and pro-
perties detected. The modifications and corrections thus sug-
gested, as necessary to be introduced into our former conclusions,
are applied, and a second approximation, but still only an ap-
proximation to truth is made, For, in investigating the laws
which regulate the several effects just mentioned, we are com-
pelled to proceed upon a new group of false suppositions. To
determine the laws which regulate the friction of surfaces, it is
necessary to assume that every part of the surfaces of contact is
uniformly rough; that the solid parts which are imperfectly
rigid, and the cords which are imperfectly flexible, are con-
stituted throughout their entire dimensions of a uniform material ;
so that the imperfection does not prevail more in one part than
another. Thus all irregularity is left out of account, and a
general average of the effects taken. It is obvious therefore,
that by these means we have still failed in obtaining a? result
exactly conformable to the real state of things : but it is equally
obvious, that we have obtained one much more conformable to
that state than had been previously accomplished, and suffi-
ciently near it for most practical purposes.
" This apparent imperfection in our instruments and powers
of investigation, is not peculiar to Mechanics ; it pervades all
departments of natural science. In Astronomy, the motions of
the celestial bodies, and their various changes and appearances,
as developed by theory, assisted by observation and experience,
are only approximations to the real motions and appearances
which take place in nature. It is true that these approximations
are susceptible of almost unlimited accuracy ; but still they are,
and ever will continue to be, only approximations. Optics,
and all other branches of natural science, are liable to the same
observations*."
• Captain Kater's Treatise on Machines.
THE LEVEE. 121
I. On the, Lever,
189. Def. a Lei)&r is a rigid rod straight or bent,
moveable in a certain plane about one of its points, which is
fixed and called its ful6rum.'
190. In a lever iohen there is equilibrium the power and
weight are to each other inversely as the perj^endiculars from the
fulorum upon the directions in which they act.
(Both the power and weight are supposed to act in the plane
in which the lever is moveable, which is technically called the
plane of the leVer).
Let AB (figs. 40, 42) or AO (fig. 41) Or BO (fig. 43), be
a lever whose fulcrum is 0; A, B the points at which the
power P and Weight TF'act; €Y, C^ perpendiculars from
upon their directions. Then the equilibrium will not be dis-
turbed by applying at two for6es F, F" parallel and equal
to P, and two others W', W" parallel and equal to W: We
haVe thus, six forces acting on the lever, of which (P, P") and
( W, W") form two couples, and the two remaining forces P',
W being counterbalanced by the reaction of the fulcrum, may
be retnoved. Hence the couple (P, P") whose arm is OY,
balances the couple {W, W") whdse arm is G^j consequently
their moments must be equal ;
.-. P. 6T= W. OZ.
191. To find the pressure on the falcrum C.
We liave shewn that P and W are equivalent to two forces
P, W acting at 0, and two equal couples (P, P"), {W, W");
these couples may be removed because they are equal and
opposite and therefore balance each other. It appears, then,
that Pand PTare equivalent to P' and W' acting at 0. Con-
sequently the pTessure on the fulcrum is the same as if
the powet and Weight Were both transposed tb it parallel to
themselves.
192. We have considered the weight of the lever incon-
siderable when compared with P and W, but if this should not
E. s. 16
122 .THE LEVEE.
be the case, let w be its weight, G its centre of gravity. Then
we may suppose the whole force w, which gravity exerts upon
the lever,. to be applied at O; this force may be converted into
a couple whose moment is w . GG, and as^ there is equilibrium
between the three couples, the sum of the moments of the two
which act in one direction [i.e. positive or negative) must be
equal that of the third ;
.-. P.GY + w.CG=W.GZ,
is the equation of equilibrium in this case.
193. Exaniples of levers of the same kind as the one in
fig. 40, are the common balance, steelyards, pokers, &c. ; and
scissors, pincers, &c. are instances of two such levers having
a common fulcrum.
Examples of levers of the same kind, as those in figs. 41j
43, are the oars and rudders of boats, cutting-knives moveable
about one end, &c.'; and tongs, sheep-shears, &c. are instances
of the combination of two such levers with a common fulcrum.
Examples of the bent lever, in fig. 42, are gavelocks, jemmies,
bones of all animals, &c.
194. We have defined a lever to be a rigid rod, but
we may consider any rigid body having a fixed axis as a
compound lever, whose fulcrum is the axis ; and if powers
Pj, Pjj, Pg...P„, act upon this lever, and balance the weights
W„W„W, Tn,, then
P,Pi + P,p^+ +P„Pn= W^w^+ W^w^+ + W^w^,
or t{Pp) = t{Ww);
the powers and weights being supposed to act in planes at
right angles to the axis, andpj, p^...p„; w^, w^...w^ being the
respective perpendiculars from the axis upon the directions in
which the powers and weights act.
This may be proved as before, by converting the powers and
weights into couples, and then transposing them into one plane ;
and it will also appear, that the pressure on the axis or fulcrum
is the same as it would be if all the forces were transported in
their own planes parallel to themselves to the axis.
THE PULLEY. 123
II. On the Pulley,
195. Def. a Pulley is a wheel of wood or metal, turning
on an axis through its centre at right angles to its plane, and
usually enclosed in a frame or case, called its hhck, which ad-
mits a rope to pass freely over the circumference of the pulley,
in which there is usually a groove to receive it and prevent its
slipping out. The pulley is said to be fixed or moveaile, accord-
ing as its axis is stationary or not. An assemblage of several
pulleys is called a system of pulleys.
196. It will be necessary before investigating the properties
of the pulley to premise, that if a cord be stretched by two equal
forces applied at its extremities in contrary directions, there will
be a tendency to break ; the force which the rope, in consequence
of the cohesion of its particles, exerts to resist this tendency,
must be equal and opposite to that which causes the tendency ;
it is called the tension of the rope. Hence tension is a force
which is exerted equally in every part, tends fr-om the ex-
tremities of a cord towards the middle, and is always equal to
either of the equal forces, by which the cord is stretched. If
one end of the cord, instead of being acted on by a force, be
fastened to a fixed point, the tension will not be altered; for
the fixed point will, by its reaction, exactly supply' the place
of the force.
1 97. In the single fixed pulley when there is equilibrium
the power and weight are equal..
Let ABK (fig. 44) be the pulley, C its centre, CN its
block ; P and W the power and weight acting at the extremities
of the cord passing over the pulley, and having the part AB'ia
contact with it. Then we may consider the pulley ABK as a
lever whose fulcrum is C; and therefore drawing the radii OA,
CB to the points A and B, we have
P.GA= W.CB;
.-. P= W.
124 THE PULLEY.
Hence it appears that no mechanical advantage is gained
by the use of this pulley ; the only purpose for which it is used
is to change the direction in which a force is transmitted.
198. To det$rmvne the jaressure on the fulcrum C.
Transpose the forces P and W to that point, and put 9
for the angle at which APaaA. BW &t& inclined to each other,
and let R be the pressure, which is, of course, the resultant
of these transposed forces, and bisects the aiigle between liiem ;
hence resolving these forces io the direction of B, we find
6
Ii = P cos - + Pcos-
- 2Pcos - .
This pressure is transmitted to N, the fixed point to which
the block is attached.
199. In the single nwveable pulley when there is eguilibrmm
the power is to the weight :: 1 : 2 x cosine of half the angle be-
tween the strings.
Let the power P act at the extremity P of the cord PABD
(fig. 45), which passing under the pulley has the part AB in
contact with it; and its other extremity fastened at D. The
weight W hangs from the block at N.
Exactly as in the last case, we find the pressure on the centre
C to be
2Pcos|,
6 being the angle between the strings AP, BD ; this force is
transmitted through the block in the direction C^, bisecting
the angle d ; wherefore the action of W must be equal to it
and in the opposite direction, otherwise there cannot be an
equiUbrinm ;
.-. PF=2Pcos|,
and consequently P : TF :: 1 : 2 cos - .
THE PULLKY. 125
200. No mechanical adeantage can be gained by the use of
this pulley, unless
d
2 cos - > 1,
Q
and .'. cos - > ^ > cos 60°;
that is, unless the strings are inclined to each other at a less
angle than 120°.
The greatest possible advantage will be gained when the
Q
strings are parallel, for then ^ = 0, and cos - = 1,
and therefore W = 2P.
201. If the weight of the pulley and its block be con-
siderable, it must be considered as an additional weight, and
added to W in the above expressions.
202. To find the conditions of equilibrium in a system of
pulleys, where each pulley hangs hy a separate string, the strings
being all parallel.
Let A^, A^, As, ••• (fig- 46) be the pulleys; M^, M^, M^ ...
the points where the strings are fastened to an immoyeable
block. Then P is the tension of the string passing under A^.
The two strings A^P, A^M^ have to support the tension of
N^Aj^; so N^A^ and M^^ support that of N^A^, and so on;
therefore,
(P=) tension of A^P : tension of N^A^ :: 1 : 2,
tension of N^A^ : tension of N^^ "1:2,
tension of ^y4, : tension of iV^gTr(== PF) :: 1 : 2;
.-. P : W :: 1 x 1 x 1 x : 2x2x2..
If n be the number of moveable pulleys, then
P ; W:: r : 2";
,-. W=^TP.
126 THE PULLEY.
203. If the weights of the pulleys and blocks are con-
siderable, let A^, A^, A^... represent the weights of the' pulleys
and blocks denoted by those letters in the figure; and let
Tj, Tj ... be the tensions of the strings N^A^, N^A^ .... Then,
as before, the weights of the pulleys must be added to the
tensions of the respective cords which they support ;
.-. P: T^+A^ :: 1 : 2;
.-. T^ = 2P-A^.
Similarly, T^ = '2T^-A^
=^2'P~2A,-A„
= 2»P-2Mj-2^,-^„
and so on, the law being manifest ; then, since the tension of
the last string = W, we have
W= 2"P- 2"-' A, - r-^A, - 2"-M, - -A,.
It appears from this expression, that the weights of the
pulleys diminish the advantage of this system.
204. If all the pulleys are equal, then
W= TP~ A^ (2"-' +2"-"+ + 1)
9"— 1
= 2"P-(2"-l)^,
= 2»(P-^.)+A;
.'. Tr-^i=2"(P-^,).
Hence, if we suppose both the power and weight diminished
by the weight of a pulley, we may then neglect the consideration
of the heaviness of the pulleys.
205. In the system (fig. 47) where each string is attached
to the weight, let T^, J'^-.-be the tensions of the first, second
...strings; then if the Weights of the pulleys are inconsiderable,
we have
THE PULLEY. 127
2;=22; = 2'P,
y,= 22; = 2'P;
and if there be n sepaxate strings,
Now W is supported by the tensions of the n strings fastened
to the block B, and
.-. W=T,+ T^+ + T„
= P(l+2 + 2''+...2"-')
2-1
= P(2''-1).
206. In the system (fig. 48), let T^, T^... be the tensions of
the first, second... strings; then Tj^ = F; and T^ has to support
three tensions equal to P; therefore
T, = P,
2; = 3?;=3'P;
and if there be («) difierent strings, the tension of the last is
?;=3''-'P.
Now the weight W is supported by two strings whose
tensions are T^, two of which the tensions are T^, &c.;
.-. W=2T^'+2T^+ + 2y„
= 2P. (1-1-3 + 3'+ 3"-')
= 2P.^
3-1
= P.(3»-1).
Remark. If the weights of the pulleys and blocks are not
■inconsiderable, they may be taken into account, in this and every
128 THE WHEEL AND AXLE.
other system, by adding each to the tension of that string which
supports it, as in Art. 203.
207. In the system, fig. 49, the weight W is supported
by the tensions of all the strings at the lower block, and as
it is the same string which passes round all the pulleys, the
tension of every part =vP; wherefore, if there be n pulleys in
the lower block, there are 2ra strings supporting the weight, and
therefore
W= 2nP.
III. On the Wheel and Axle.
208. The wheel and axle consists of a cylinder and a wheel
firmly attached to each other, and being moveable about a fixed
axis coinciding with the axis of the cylinder, and passing through
the centre of the wheel at right angles to its plane, as in fig. 50.
The power P acts by means of a cord wrapped round the cir-
cumference of tte wheel O, and the weight W is fastened to a
cord which is wound upon the cylinder AS as P turns the
machine round its axis ; and thus Wia raised.
209. To find the condition of equilibrium on the wheel and
axle.
We may consider P and W as forces acting upon a rigid body
with a fixed axis, and therefore their moments about that axis
must be equal ;
.". Px (perpendicular upon its direction firom the axis),
= W. (perpendicular upon its direction from the axis).
Now these perpendiculars are respectively the radii of the
wheel and of the cylinder ;
.•. P. (radius of the wheel) = W. (radius of the axle).
210. If the thickness of the rope be considerable, it must be
taken into account.
We may suppose the actions of P and TF to be trans-
mitted along the middle or axis of the rope, and then the per-
THE INCLINED PLANE. 129
pendiculars upon the directions of F and W will be respectively
equal to
radius of wheel + radius of rope,
and radius of axle + radius of rope,
and the condition of equilihrium is
P . (rad. wheel + rad. of rope) = W (rad. axle + rad. of rope) .
This diminishes the advantage of the machine.
211. The pressure on the axis of this machine may be found
by transposing P and W in their own planes, parallel to them-
selves, to the axis.
IV. On the Inclined Plane.
212. This machine is nothing more than a plane inclined to
the horizon. The condition of equilibrium may be thus found.
Let A^ (fig. 51) be the plane ; A parallel and BC perpen-
dicular to the horizon ; W the weight, P the power. Draw TFS
perpendicular to the plane, WG perpendicular to the horizon.
P is supposed to act in the plane R WB. The weight W is kept
at rest by three forces, viz. P in the direction WP: gravity (= W)
in the direction WG, and reaction R of the plane in the direc-
tion WR.
Denote the angle PWB by 6, and the inclination BA G of
the plane to the horizon by i; and resolve the three forces,
acting on the point W, in a direction parallel to the planes;
the sum will be
PcobPWB- W cos AWG + B cos RWB
= Pcos 9 — TF. sin i.
But since there is an equilibrium, this sum must be equal to
zero,
.•. Pcos 5= Wsini,
which is the condition of equilibrium.
E.8. 17
130
THE SCREW.
213. If P's direction should happen to be parallel to the
plane, 6 = and cos ^ = 1 ;
.-. F= Wsinl
But if P's direction should happen to be parallel to the
horizon, d = — i and cos (— i) = cos i;
.". Pcosz= TFsint;
/. P= Wta.ni.
214. To Jind the reaction of the plane.
Kesolve the forces in a direction at right angles to that in
which P acts :
.-. B sin BWF- Wain G WF= 0,
or Bcoad-Wam{90 + {+6)=0;
cos a
V. On the Screw.'
215. This mechanical power is a combination of the lever
and inclined plane; it may be conceived to be thus gene-
rated.
Let ABCD (fig. 52) be a cylinder; BEFO a rectangle
whose base BE is equal to the circumference of the cylinder.
Divide this rectangle into any convenient number of equal
rectangles QFJ, IH, CK; and draw their diagonals BH, OK,
IF. Then, if this rectangle CE be wrapped upon the cylinder,
so that BE coincides with the circumference of the base, E, H,
K, F will respectively fall upon the points B, G, I, of the
cylinder, and the lines BH, GK, IF will trace out upon its
•surface a continuous spiral thread BLGMINO winding uni-
formly up the cylinder. The cylinder is usually made pro-
tuberant where the spiral line BL GMINC ialh upon it so that
the thread becomes a winding inclined plane, projecting from
the cylinder as in fig. 53, and differing from the inclined plane
THE SCREW. 131
BH* in nothing but its winding course. This is the external
screw. The internal screw is formed by applying the paral-
lelogram BEFO to a hollow cylinder, equal to the former, and
making a groove where the thread falls to fit the protuberant
thread of the external screw. This internal screw is often called
a nut, and the other the screw. When the two screws are thus
adapted to each other, the external or the internal screw, as the
case requii'es, may be moved by means of a lever about their
common axis, as in figs. 54, 55. The force being applied to
the lever at right angles to jt, in a plane parallel to the base of
the cylinder.
The screw and nut thus applied to each other, resemble two
inclined planes, such as BHQ and HBE, one of which is laid
upon, and slides down the other ; and as the planes wind round
the cylinder a rotatory motion ensues. When the machine is
worked, the weight is laid upon the nut, and thus causes its
inclined plane to press upon that of the screw in the direction
of gravity. The consequence would be, that the nut and weight
with it would begin to slide down the thread of the screw and
descend, but this is prevented by confining the nut so that it
cannot have a rotatory motion, but only one of ascent or descent.
The screw is then turned round by means of a lever passing
through its head, and thus its inclined thread sliding under that
of the nut, forces the nut and the weight upon it to ascend, just
as by pushing the inclined plane EBH in the direction EB, the
• The following illustration renders this very clear : —
*' When a road directly ascends the side of a hill, it is to be considered as an in-
clined plane ; but it will not lose this mechanical character, if, instead of directly
ascending towards the top of the hill, it winds successively round it, and gradually
ascends so as after revolutions to reach the top. In the same manner a path may be
conceived to surround a pillar by which the ascent may be facilitated upon the prin-
ciple of the inclined plane. Winding stairs constructed in the interior of great
columns partake of this charaeler; for although the ascent be produced by successive
steps, yet if a floor could be made sufficiently roush to prevent the feet from slipping,
the ascent would be accomplished with equal facility. In such a case the winding
path would be equivalent to an inclined plane, hent into such a form as to accom-
modate it to the peculiar circumstances in which it would be required to be used.
It will not be difficult to trace the resemblance between such an adaptation of the
inclined plane and the appearances presented by the thread of the screw; and it may
hence be easily understood that a screw is nothing more than an inclined plane,
constrncted upon the surface of a cylinder."— Captain Kateb's Machines.
132 THE SCREW.
plane QBH would be made to ascend. One turn oif the screw
raises the weight through an altitude equal to the distance be-
tween two threads. Sometimes, however, the nut is firmly fixed
so as to admit of no motion whatever (as. in fig. 54) ; and then
the thread of the screw, in sliding under that of the nut, forces
the screw to descend and press violently against any obstacle
which may be opposed to it. In some cases the weight is not
applied to the nut, but to the screw ; but as the two inclined
planes are perfectly equal and similar, it will require the same
force to support a weight on one ^s on the other, and for this
reason one investigation will serve for both.
As before observed, the screw is worked by applying a
power P at the end of a lever ; and the moment of P to turn
the screw round
= P X length of the lever,
and therefore P is equivalent to a force
P X length of the lever
rad. cylinder
acting immediately at the thread of the screw in a horizontal
direction parallel to that in which P acts. Now the inclined
plane on which W rests, by means of the nut, is only BH
wrapped round the cylinder; its inclination to the horizon or
base of the cylinder is therefore HBE.
Hence we have
^^ length of lever ^^,^^^ (Art. 213)
rad. 01 cylmder
^ BE
_ T^r distance between two threads
~ ' circumf. of cylinder
But the radii of circles are proportional to their circum-
ferences ;
length of lever _ circumf. described by power
rad. of cylinder circumf. of cylinder '
THE WEDGE. • 133
p circumf. by power _ tt?. dist, between two threads _
' circumf. of cylinder ~" ' circumf. of cylinder '
„ „7- dist. between two threads
circumf. described by power '
As the distance between two successive threads can be made
very small, and the circumference described by the power as
large as we please, the advantage of this machine is very great ;
and it is remarkable, that it does not depend upon the thickness
of the screw.
VI. On the Wedge.
216. A wedge is the solid figure defined by Euclid (Book
XII. Def. 4) as a. triangular prism. Its two ends are equal and
similar triangles, and its three sides rectangular parallelograms
(see fig. 56). It is principally used in splitting timber, and
separating bodies which are very strongly united, and in raising
very heavy weights through a small altitude, for the purpose
of introducing a lever, or some other more convenient machine.
AB is called its edge, GDEF its head, GABD and FABE its
faces.
When used, its edge is introduced into a small cleft pre-
pared to receive it, and then by violent blows with a hammer
on its head its body is driven between the substances, which
are thus separated by an interval equal to the breadth of the
head. After this, a larger wedge may be introduced, if neces-
sary, and treated as before, until the requisite degree of sepa-
ration is effected.
As the wedge is di-iven in by violent blows, if its sides
were perfectly smooth it would start back by the pressure of
the obstacles upon them in the interval between the strokes;
and thus we should fail in effecting and maintaining the requi-
site degree of separation, and the machine would be rendered
useless. In practice, however, the friction in this machine is
always so great as to prevent any recoil, and forms, in fact, the
principal resistance to be overcome in driving the wedge. The
mode of working this machine will at once present itself to the
134 - GENERAL PROPERTY OF MACHINES.
reader as Ibeing totally different in principle from that of all the
other machines we have described. These are made to work
by the constant and steady exertion of a power, uniformly press-
ing upon that point of the machine at which it is applied, and
gradually producing motion in the weight ; but in this machine
motion is accumulated in a hammer, by suffering it to descend
fi'om an altitude, and is suddenly by an impulse transferred to
the wedge. In this case it must evidently be a useless labour
to attempt to calculate the ratio of P to W, when they act by
pressures, as in the other mechanical powers, and are in equi-
librium. It is true, when we know this ratio, a slight increase*
of P will gradually produce a motion in W, and thus separate
the obstacles; but this mode of working the machine is so
widely different from that actually practised, that it would be
a waste of time and labour to attempt an explication on Statical
principles. A slight^stroke with a hammer is found to be far
more effective than several tons of pressure. The only theo-
retical property of the wedge which agrees with practice is that
its advantage is increased by diminishing its angle DBE.
All cutting instruments, such as knives, swords, hatchets,
chisels, planes used by carpenters, nails, pins, needles, &c. are
modifications of the wedge. Of these, knives, planes, pins and
needles, are usually worked by pressure, but swords, hatchets,
chisels, nails, &c. are worked by percussion.
GENEEAL PROPERTY OF MACHINES.
217. If the nature of a machine be such, that when the
power and weight balance each other in one position of the
machine they will balance in every position of it, a very re-
markable property appertains to it, deducible from the principle
of virtual velocities, which we may state as follows :
The power is to the weight as the space moved through iy the
weight when the machine is put in motion is to the space moved
* This, however, supposes the sides to be perfectly smooth, for otherwise the
friction itself, without the assistance of any power at all, would preserve the equili-
brium.
white's pulley. 135
through hy the power in the same time ; the spaces being measured
respectively in the directions in which the power and weight act.
Let tlie whole space (measured thus) through which the
power P moves be divided into a very large number of spaces
Sj, Sj..., and let s\, s',... be the corresponding spaces described
by the weight W; then
S=Si + «j+
t 1,1,
s=s, + s^+
But because P and W are always in a position of equilibrium ;
«i, s\, are their virtual velocities for the first position;
.-. Ps^+Ws\ = 0,
Similarly Ps^ +'Ws'.^ = 0, for the 2nd position
Ps^ + Ws'^ = 0, 3rd
.-. Ps+Ws' = 0;
P __£
•'■ W s'
This equation expresses the property enunciated. The negative
sign points to the fact, that the direction of the action of one of
the two forces P, W is opposed to the direction in which the
point moves on which it acts.
Mechanical powers possessing this property are ; —
(1) The straight lever supporting weights.
(2) Air the pulleys in which the strings are parallel. '
(3) The Wheel and Axle.
(4) The Screw.
(5) The Inclined Plane, only when the Power haaigs by a
string passing over the top of the plane.
WHITE'S PULLEY.
218. In the common systems of pulleys each pulley has its
own independent centre of motion; and consequently as they
136 hunter's screw.
air move with different velocities and with different degrees of-
pressure, some of them will be liable to greater wear than others,
which will very much tend to increase the friction and other
inequalities and resistances ; and will greatly diminish the
efficiency of the machine. To obviate these difficulties, Mr
James White invented a system of pulleys (fig. 57), consisting
of two blocks A, B, into which grooves were cut, the radii of
those in the upper block being as the numbers 1, 3, 5... and the
radii of those in the lower block being as the numbers 2, 4, 6...
Now, suppose the lower block to be raised through one inch,
then each of its strings will be shortened one inch, and therefore
the circumference of the pulley BB^ describes one inch ; that of
AA^^, two inches; that of BB^, three inches, and so on; which
numbers being proportional to the radii of the respective pulleys,
they will all move, with the same angular velocity; and, con-
sequently, each block instead of being composed of separate
pulleys may consist of one solid piece of wood or metal, contain-
ing the grooves before mentioned. The disadvantage of this
system is, that if the cord be at all elastic it cannot be kept
stretched in every part on account of the tension not being the
same throughout, so that the smaller grooves are rather a hin-
drance to the motion than a help.
HUNTER'S SCREW.
219. We have seen (Art. 215) that the advantage of a screw
increases in proportion as the distance between the threads dimi-
nishes, and as the length of the lever at which the power acts
increases ; therefore, by making the threads of the screw suffi-
ciently fine, we may increase the advantage as much as we
please ; but there is a limit to the fineness of the threads ; for as
all the weight is borne upon them, if they are too fine they will
not be sufficiently strong to bear the load. If we, on the other
hand, increase the length of the arm of the lever, with the view
of increasing the advantage of the screw, the power will have to
describe an inconveniently large circle. To obviate these natural
defects, and yet increase the advantage to any degree, Mr Hunter
invented the screw in fig. 58 ; A and B are two common screws,
of which A is also a hollow screw to admit B, which is fastened
hunter's screw. 137
to the moveable plate D of wpod or metal. If D, d be the dis-
tances between two threads of the screws. A, B respectively;
then, while the power describes one circumference, A descends
through Z>, and B ascends in A through d, and the space de-
scended by the plane D\^D—d; for when A descends it carries
B along with it, though B is at the same time ascending in A.
Wherefore, by Art. 217,
P. (circumf. described by P) = PF. (-D -^ <?) ;
W _ circumf. described by P
•"• P ^^^ •
Now we can make D and d as nearly equal as we please
without diminishing the strength of the machine, and therefore
the advantage of this screw admits of indefinite increase.
220. It appears from Art. 209, that the advantage of a
wheel and axle is
rad. of wheel
rad. of axle '
which might theoretically be augmented ad lilitum, either by
increasing the radius of the wheel, or by diminishing that of the
axle. But by the former means, the power would practically
have to describe an inconveniently large space, and the machine
would become cumbrous; and, in the latter case, it would be
too weak to bear the pressure of the weight upon its axle. To
remedy these inconveniences, and at the same time to increase
the advantage in any requisite degree, the form of fig. 59 has
been given to it ; where A is the wheel, B and two axks of
unequal radii, firmly fixed to each other, and having the same
axis. The cord BDC as P descends is wound upon the axle B
with the larger radius, and is at the same time unwound from
the axle G with the smaller radius ; it passes imder a pulley D,
to which the weight Wis, attached. Let R be the radius of the
wheel, rr' those of the axles B, G. Then when the machine
turns once round, P descends through 2vR, and the length of
the cord wound upon B is 27rr, and the length unwound at the
same time from G is 27r/ ; wherefore, upon the whole, the length
of cord hanging down from the axles is diminished by
27rr - 27rr' ;
E. s. 18
138 THE GENOU.
and, therefore, W has ascended through
Trr — Trr'.
Wherefore, by Art. 217,
P : W :: -irr-irr' : ^irB,
:: r- r : 2^;
.•. the advantage = ^ = t-
As we can diminish the denominator of this fraction as much
as we please, without weakening the materials of the machine,
there is no limit to the advantage of it, except what arises from
the very great length of cord that must he used in raising W
through a given space.
THE GENOU.
221. This instrument is represented in its simplest form in
fig. 60, where AF is the profile of a frame in which the rods
AB, BC work. AB is moveable about a fixed axis passing
through A; it is connected with BC by a compass joint at B;
and the other end of BC, by means of a pin passing through
it, is compelled to move in the vertical groove EF. The
power is applied at G, a point in AB, in the plane of the rods
ABO. I It causes B to come nearer to AF; and, consequently,
G presses downwards upon any obstacle opposed to it. It is
obvious this machine is only applicable in those cases in which
G is required to descend through a small space, as in printing,
where it presses the paper upon the type.
Let W=the reaction at C, P the power applied horizon-
tally at G^, e = the angle BAF, a = AB, h = BG, c = AG, and
let GP intersect AF in p. Then Gp = c sin 6, and therefore the
virtual velocity of P
= d{Gp) = ccose.dd.
Also AF= acoBd + h cos BCA, therefore the virtual velocity
= d {AF) = - a sin ddO - b sin BCA . d {BCA).
WHEELS AND AXLES. 139
Now sin BGA = r sin ;
and .-. cos BGA . d [BGA) = | cos ddd;
.'. the virtxial velocity of W
• a -in • n O' COS QdO
= — asm a . da — a sm a . T . ^^ .
I cos BGA
/a COS0 N • a JO
= - 1 + T . d7T7 . a sin . d0.
V COS -604/
Wherefore, by Art. 114, the advantage of the machine
^w COS ede
P~ ( a cos6 \ . „j.
1 + T • tttth . a sm odd
\ b cos BGAj
_ c h cos BGA . a cos 6
a ■ a sin 5 (a cos + & cos BGA)
_ AG.Cb.hA
AB.Bb.AG'
where Bb is drawn parallel to GF.
222. A combination of wheels and axles may be used in-
stead of the machine in Art. 220, when that is inconvenient and
great advantage is required. Fig. 61 represents a combination
of three of these mechanical powers. An endless strap passes
over the axle a and the wheel B, and another strap passes over
the axle b and the wheel G, If two successive wheels are
required to turn in opposite directions, the strap must be crossed
as between A and A in the figure ; when the wheels are to turn
in the same direction, the strap must not be crossed. B and G
are turned by the friction of the straps upon their surfaces ; and
hence it is manifest, that if the force to be overcome by any
wheel be greater than the friction of its strap, the strap will slip
round without carrying the wheel with it, and the action of the
machine will cease. Wherefore, in order to make the friction
upon the surfaces of the wheels and axles as great as possible,
ihey are covered with leather, which is nailed or glued on
140 TOOTHED WHEELS.
them ; and both this leather and the concave sides of the straps
are suffered to be in a rough state ; the friction is also increased
by crossing the straps.
To calculate the advantage of this combination, denote the
tension of the strings d and e by T, T ; then since P balances
the tension J' on the axle a, we have, by Art. 209,
T _ rad. of wheel A
P rad. of axle a
o- -1 1 T' rad. of wheel B
Similarly, y= ,ad.ofaxle& '
and
W rad. of wheel O
T' rad. of axle c '
and, therefore, by multiplying these equations together, we have
W _ product of radii of all the wheels
P product of radii of all the axles '
TOOTHED WHEELS.
223. By far the most general modification under which
wheels and axles are used in practical Mechanics, is that of
toothed wheels.
Let A, a (fig. 62) be the centres of two wheels BG, he, upon
the circumferences of which let teeth or cogs D, E, F, d, e, f,
of any proposed form, be raised at equal distances all round ;
in order that this may be possible, the radii of the two wheels
must be in proportion to the number of teeth that are to be
constructed upon them. If one of the wheels (be for instance)
be turned round its axis a, its teeth will press upon the teeth
of the other wheel BO, and turn it round its axis ^ in a con-
trary direction, and as two corresponding teeth F, f separate
from each other in consequence of the motion, two others D, d
come in contact ; and thus the wheel a is enabled to produce a
continuous motion in the wheel A. Similar teeth are con-
structed upon the axles of each wheel, and the axle so prepared
is called a pinion, and its teeth are called leaves. From the
nature of the wheel and axle it is manifest that motion is com-
TOOTHED WHEELS. 141
municated to each wheel, in this modification, by a pinion in
which it runs as in fig. 63, where P descending turns with it the
pinion a which turns the wheel B, and this carries with it the
the pinion h which turns the wheel and axle c, and raises the
weight W. In this case, as in Art. 222, it is clear that
W _ product of the radii of the wheels
P product of the radii of the axles
_ radius of A product of number of teeth in the wheels
radius of c product of number of leaves in the pinions '
Here there are no teeth ia A and c, on which account we have
not reduced their radii to equivalent numbers of teeth.
224. In the description of toothed wheels we have said
that the teeth or cogs are to be of any proposed form, because
in fact they are commonly made in any form that meets the
fancy of the maker. It must not be imagined, however, that
all forms are equally advantageous, as we shall easily tmder-
stand by referring to fig. 62, and tracing the actions of the
teeth upon each other during their motion. Suppose lo to begin
to turn round, and let us trace the actions of d and D. When d
first comes in contact with D, the latter presses against the side
of eZ in a single line of points, very near the extremity of d, in
the direction of a normal to the side of d, that is, in the direction
jpD perpendicular to the radius ad. Therefore, drawing Ap
parallel to ad, the action of d may be transmitted to p, and its
eflSciency varies as Ap. But as the wheel he continues turning,
the point of contact D slides along the side of d, and thus pro-
duces a very strong friction, and consequently rapid wear both of
the side of d and of the edge of the tooth D. This goes on
till d and D come into the position e and E, when their sides are
for a moment in contact, and then the efiiciency of d in turning
D varies as AD.
When the teeth d and D leave this position a similar action
to what has just been described commences, only it is in a
reverse order ; and the edge of the tooth d presses against and
rubs the side of the tooth D.
It appears then, with teeth of the form of those in this
figure, —
142 TOOTHED WHEELS.
1st. That the efficiency of the pressure which one tooth
exerts upon another, and consequently the motion "produced, is
very irregular, being in one position proportional to Ap, and in
another to AD.
2ndly. That the edges of the teeth are subject to very rapid
wear in consequence of ruhhing with a single line of points in
contact with the sides of the teeth of the other wheel, which
latter is thereby also very soon worn hollow, and the whole
rendered useless.
3rdly. That in consequence of the rubbing of the teeth
against each other much of the power is rendered ineffective.
4thly. That since there are favourable and unfavourable
positions, the power must be sufficient to move the weight in
the most unfavourable position with the requisite degree of
celerity; and consequently, when the machine is in the most
favourable position there will be an excess of power which will
cause the machine to move much too rapidly, and often produce
fractures; nothing in fact having so great a tendency to tear
asunder the parts of a machine and render it useless as an
irregular motion of this kind.
From these considerations it will at once be evident that the
best form of the teeth will be, when, —
1st. The teeth of one wheel press upon those of the other
in such a direction that the efficacy may be uniform; that is,
such that the perpendiculars upon that direction from A and a
are of constant lengths.
2ndly. The teeth of one wheel do not rvh but roll upon
those of the other.
Srdly. The motion of one tooth upon another is uniform.
When these conditions are fulfilled, it is also necessary that
the distances of the axes of the wheels should be such that as
great a number of teeth may be in contact at one time as pos-
sible, and that there may be no jolting nor violence of any kind
TOOTHED WHEELS. 143
when two teeth separate or come in contact. These precautions
will very much diminish the chances of fracture.
Many forms of teeth have heen proposed fulfilling one or
more of those conditions, hut it seems to he agreed on that the
following is the best. ,
225. Let ABB (fig. 64) be a given wheel on which it is
proposed to erect teeth ; and let AB be the proposed breadth
of a tooth. Upon AD wrap a string and fasten it at D. Then
unwrap it, beginning at A, and its 'extremity A will trace out
the curve Aa called the involute of the circle AD. In a similar
manner, describe the involute Bb intersecting the former in C;
then A GB will be the tooth required, which may be taken as
a pattern of all the others to be formed upon the wheel. In a
similar manner the leaves of the pinion may be found, by first
constructing a pattern by means of the involute of its circum-
ference. Let PL be a position of the thread whose extremity
generates the involute Aa ; then we may suppose the point L
to be fixed for an instant^ and therefore P will begin to describe
an arc of a circle whose centre is L, and therefore PL is a normal
to the curve A 0, and OL the perpendicular upon this normal is
constant. In the same manner it may be shewn, that the per-
pendiculars upon the normals to the leaves of the pinion are all
constant and equal to the radius of the pinion. Wherefore, since
the leaves of the pinion press against the teeth of the wheel in
the directions of normals at the points of contact, and the per-
pendiculars on these directions are always of the same length,
the action will be uniform, and consequently the motion will be
uniform also.
THE ENDLESS SCREW.
226. This machine, represented in fig. 65, consists of a
screw A whose axis is 5(7; and a wheel and axle D, E; th©
wheel being furnished with teeth exactly fitting the threads of
the screw. The screw is turned by means of the winch CP,
and its thread instead of pressing against a nut, presses against
the teeth of the wheel, and forces them forward ; each turn of
the screw or winch, advancing the wheel one thread of the
144
TOOTHED WHEELS.
screw; or, which is the same, one tooth of the wheel. The
winch must therefore he turned round as many times as there
are teeth in the wheel, in order to turn the axle E once round.
Wherefore, putting B for the radius of the circle described by
the power P; r for that of the axle E, and n for the number
of teeth in the wheel D ; the circumference described by P
and therefore the space described in one turn of the wheel D
= 2mrli.
But the space ascended by PF in the same time
= the circumference of the axle E
= 27rr.
Consequently, by Art. 214,
W 2mrR R
P ^irr r
ON BALANCES.
227. A halance is any instrument invented for the purpose
of comparing the heaviness of different bodies ; that is, for ascer-
taining their weights.
The common halance (fig. 66) consists of an inflexible rod
AB, called the beam, resting upon a fulcrum Cat its middle
point ; from its extremities A, B are suspended two equal scales
D, E by means of fine chains or strings. The fulcrum O and
the points of support are in the same straight line, but the
centre of gravity of the beam is a little below C. In this state
the balance when unloaded ought to rest with its T)eam AB in
a horizontal position. If a weight be put into one of the scales,
the common centre of gravity of the scale and its load will be
in the vertical passing through the point of support (Art. 131) ;
and therefore we may transmit both the scale and its load to the
point of support. Wherefore, when weights are placed in the
scales, we may suppose them placed immediately at A and B,
and therefore the balance becomes a straight lever whose ful-
crum is G; and since the arms AC, BG are equal, there will
be an equilibrium when the weights are equal (Art. 190). If the
THE COMMON BALANCE. 145
weights are unequal, let G (fig. 67) be the centre of gravity of
the beam AB in the oblique position assumed in consequence
of the inequality of the ■weights. Let w be the weight of the
beam, which by Art. 130 we may suppose to be "placed at O;
8 the weight of each of the equal scales ; P, W the weights in
-D and S respectively ; d = the inclination of the beam to the
horizon. Then the lever is kept at rest by three parallel forces,
viz. 8+P&tA, S+ WsitB, and w at G. The perpendiculars
from upon the directions of these forces are
AGcosO, CB cose, and GOsind:
therefore, by Art. 194,
{8+P). AG COS0 + w. GOaine ={S+W). BO cos 0;
.'. P.AC+w.GCt&ne = W.BG,
by dividing by cos^, and observing that AG= BG;
^ „ W-P AC
w GO
The sensibility of a balance consists in the beam attaining
considerable obliquity, when the difference between P and W
is extremely small; and therefore the obliquity attained by
different balances when loaded with the same weights, might
be taken as a measure of their respective sensibilities. As W— P
is constant in this case, and as d is very nearly equal to tan 0,
we may use
AC
w.GC
as the measure of the sensibility.
A different measure of sensibility is however generally used,
which may be thus explained. Let B be the difference between
W and P which produces a given very minute appreciable devia-
tion 0' (which is the same for all balances) ;
.'. or tan =— . -tt-p^;
w GO
n (rO a'
E. S.
19
146 THE STEELYAED.
the ratio of the whole pressure P+ W+28+w (Art. 194) on
the falcrum to this weight is taken as the measure of the sensi-
bility, or neglecting 6' in this measure which is the same for all
balances, and using 2P+ 28+ w for the pressure on the fulcrum,
the fraction
F+S+^'w AB
w • GO
is the measure generally employed. From either of these mea-
sures we derive the following general results : —
That the sensibility of a balance is increased,
(1) By increasing the length of the beam.
(2) By diminishing the distance of its centre of gravity
from the fulcrum.
(3) By diminishing its weight.
For further information on subjects connected with the com-
mon balance, the reader is referred to Captain Kater's Treatise on
Machines.
THE STEELYAED, OR ROMAN BALANCE.
228. This instrument is a lever AB (fig. 68) with unequal
arms A C, CB ; the fulcrum being C. As it is commonly con-
structed, the longer arm A G preponderates over the shorter CB ;
let therefore G be the centre of gravity of the beam AB, at
which point we may suppose its weight w collected. And let P,
a given weight suspended from p, balance W, the body to be
weighed suspended from B. Then (Art. 194)
P.Cp + w.GG= W.GB;
P.Gp + w.CG
• GB '
xP.Cp + w. GG,
oc Gp + 'j,.GG.
Now let D be such a point that when P is suspended from D,
it just balances the beam ;
THE DANISH BALANCE. 147
.-. P.GD = w.GO;
.-. TFx Cp+CD:x Dp.
It appears therefore, that the weight W is proportional to
the distance of p from D. If when p is at E, W is one pound,
then making EF, FH, HI...ea,ch equal to DE; when p is at
F, H, I... TFwill be 2 lbs., 3 lbs., 4lbs., ... respectively, and we
may number the points E, F, H...\,2,^,... respectively ; and if
the spaces DE, EF. . . be subdivided into sixteen equal parts,
each of them will correspond to one ounce, and we shall be able
to ascertain W with corresponding accuracy by sliding the
weight P along the arm A C until it comes into such a position as
to balance W, and then reading off its place, which will be the
number of pounds and ounces which express its weight.
The practical advantage of this balance "is, that it requires-
but one weight F, and the pressure on the fulcrum, on which the
friction depends, being equal to F+ W, is less than the common
balance so long as the substance to be weighed is heavier than
F; on the contrary, however, when the substance to be weighed
is not so heavy as F, the pressure on the fiilcrum is greater than
in the common balance, and consequently the friction, which
diminishes the sensibility of the machine, is greater ; and, there-
fore, for the determination of small weights the common balance
is to be preferred, both on account of the diminution of friction
and also because small weights can be more accurately subdi-
vided than small spaces on the arm.
THE DANISH BALANCE.
229. This instrument consists of a lever AD (fig. 69), at
one end A of which is fastened a given weight A, and at the
other B a dish D to receive the substance to be weighed. The
fulcrum or point of support G is made to slide along AB until
the beam is horizoijtal, and by its place on the graduated beam
AB the weight of the substance put into the scale-pan is deter-
mined. The method of graduating the beam AD may be thus
148 kobekval's balance,
investigated. Let G be the centre of gravity of the instrument
(including the beam, weight A, and scale-pan* D), P its weight ;
TFthe weight in the scale D. Then we may suppose P aipplied
at G (Art. 133), and since there is an equilibrium between P and
PF" about the fulcrum 0,
.: W.BC=P.CG = P.XBG-BC)
= P.BG-P.BG;
P.BG
BC-.
P+W
Wherefore, if P be mlbs, and W has the values 0, 1, 2,
3 lbs.... BG has the values
n
BG n.BG n.BG n.BG
n ' n + 1 ' n + 2 ' n + 3 '"
which quantities are in harmonical progression, because their
reciprocals are in 'arithmetical progression. The spaces 0, 1 ;
1, 2 ; 2, 3 ; ... may be again subdivided, if necessary, and when
this beam is thus prepared, the weight Wmay be ascertained
with as much facility as in the common steelyard ; but the dis-
advantage of this balance is, that as the weight increases the
intervals between the divisions become smaller, and consequently
it is not so well adapted for determining large weights as small
ones.
EOBEEVAL'S BALANCE.
230. This machine consists of four stra^ht rods AB, Bh,
la, a A (fig. 70), forming a parallelogram in a vertical plane, and
being connected by compass joints at B, h, a, A; at C and D
the middle points of the rods AB and ab there are fixed axes
about which they are moveable ; GE, FH are two horizontal
rods rigidly connected with Aa and Bh, fi-om which the equal
weights P and Q are suspended. The peculiarity of this balance
is, that P and Q will be in equilibrium from whatever points
of the rods GE and FH they are suspended. To prove this
property, suppose the machine to be put in motion; then if
The acale-pan is here supposed to be transmitted to B.
robeeval's balance. 149
A ascends, B will descend through an equal space ; and as
ABha must necessarily continue to he a parallelogram, Aa and
Eb will continue parallel to CD, and therefore each vertical ;
wherefore E will ascend and F will descend through spaces
respectively equal to those descrihed by A and B, and there-
fore equal to each other. It is also manifest, since Aa and Bh
continue vertical during the motion, that OE and FH continue
horizontal, and consequently the space ascended by P is equal to
that descended by Q, wherefore they satisfy the equation of Art.
217, and are consequently in equilibrium in every position.
GHAPTEE VIII.
OK FRICTION.
231. The resistance to rotatory and progressive motion
in bodies which rub against surfaces with which they are in
contact, is zaW&di friction, and is distinguishable into two kinds.
(1) Statical friction, or resistance to the prodziction of motion
in a quiescent body.
(2) Dynamical friction, or the resistance which diminishes
existing motion.
Of these two kinds, since all machines are designed to work,
the latter is of more importance in practical Mechanics.
232. There are three ways in which one surface can move
upon another, and hence both Statical and Dynamical friction
are divided into three corresponding heads.
(1) When the surfaces in contact are two planes.
(2) When the surfaces in contact are a solid and a hollow
cylinder.
(3) When a cylinder rolls (without rubbing) upon a plane.
The laws which govern the action of friction cannot be
deduced from theoretical considerations, though theory will
render us great assistance in our researches by pointing out
the experiments which are most likely to lead us to the dis-
covery of them, as well as shewing the inconclusiveness of
other experiments, on which we might otherwise be induced
to rely. It is to be regretted, however, that the experiments
FRICTION. 151
■which have been made upon the subject by different philoso-
phers are frequently at variance ; and, consequently, the theory
cannot be said to have arrived at that state of perfection which
is desirable.
233. The statical friction of plane surfaces is, under like cir-
cumstances, proportional to the pressure.
For let AB, ab be two planes in contact, placed in a hori-
zontal position, the lower one AB being firmly fixed, but the
upper one ah free to slide upon it. To ab attach a horizontal
string ID passing over a pulley D, and having a dish G sus-
pended from it. Load ah with a weight w, and denote the
whole pressure of the plane ah on AB by W. Pour fine sand
into the dish G until it begins to move, and then the weight of
the dish and sand is the measure of the statical friction of the
planes corresponding to the pressure W. If ah be loaded with more
weights until the pressure is 2W, the friction is found to be
double of what it was before ; when the pressure is 3 IF, the
fraction is trebled; and so on. Wherefore the statical friction
of plane surfaces is proportional to the pressure.
This result was confirmed by Coulomb and Ximines for very
considerable pressures; in extreme cases, where the pressures
were very large Indeed, the friction was observed to be rather
less In proportion than for small pressures ; the deviation from
the above law was however so small, even for extreme cases,
that we shall not fall into any very considerable error In sup-
posing the law to be universally true.
The following method of establishing the property of the
proportionality of the friction to the pressure. Is very convenient
for experiments.
Let the body W (fig. 51) be placed upon an inclined plane
AB, and then let the altitude BGhe slowly Increased until the
plane has acquired such an elevation that W begins to slide
down it ; at this moment the friction just balances the weight
W, and since It acts parallel to the plane In the direction AB,
we may consider W as kept in equilibrium by a power In that
direction, hence
152
FRICTION.
friction . .,,,„,„>
" PP — =sm«, (Art. 213)
W 1
— - — =-!-., (Art. 214)
pressure cos ^ ^ '
friction sin i
. = tan^;
pressure cos i
.'. friction = (pressure) . tan i.
234. The fraction , is usually called the coefficient
pressure
of friction, and is taken as its measure. It appears then, that
in the last experiment the coefficient of friction is equal to the
tangent of the inclination of the plane.
235. It being granted that the friction is proportional to
the pressure when the surfaces are given, then, whatever be the
magnitude of the surfaces in contact, the friction will remain
the same, so long as the pressure is the same.
Let the body W (fig. 51) have faces, whose areas are C and
P square inches ; then when the first face is in contact with the
plane, the whole pressm'e is supported on square inches, and
therefore the pressure on each square inch, is equal to
pressure
and therefore the friction upon each square inch of surface
pressure ^
= *- — ry — .tant.
Consequently the friction upon the whole surface
pressure ^ . i r ■ i.
= ^ — ^ — . tan t X number oi square inches
pressure . . „
= ~ — py — .tan IX u
= (pressure) . tan i.
In the same way it may be shewn that the friction upon the
second surface
= (pressure) . tan *,
FRICTION. 153
and therefore the friction of a body is the same whether the
surface on which it rests be large or small. When the surface
is very small in proportion to the weight, the pressure on each
square inch becomes very large, aijd then the friction, as ob-
served in Art. 233, becomes somewhat less in proportion to the
pressure; and therefore the friction is less, in a slight degree,
when the body rests upon a small surface than a larger.
236. These are the chief properties of statical friction; it
does not belong to us to investigate those of dynamical friction ;
but to make the subject complete we shall annex the following
summary of results which have been obtained by various experi-
mentalists.
(1) Dynamical friction is a uniformly retarding force : and
it diminishes as the pressm^e increases.
This is only true when the surfaces in contact are hard ; for
in experiments made with bodies covered with cloth, woollen, &c.
the friction was found to increase with the velocity.
(2) In the same body Statical friction is greater than Dy-
namical friction ; i. e. it requires a greater force to put a body
at rest in motion, than is requisite to .preserve the motion un-
diminished when once it is produced.
This was thought by Professor Vince to arise from the
cohesion of the body to the plane when it is at rest, which does
not happen when the body is in motion.
(3) When a body of wood is first laid upon another, the
Statical friction increases for a few minutes, when it attains its
maximum, and no further alteration takes place. In making
experiments, therefore, it is necessary to wait some time before
the body is put in motion.
(4) Friction between substances of the same kind is greater
than when they are of different kinds.
(5) The velocity has very little, if any, influence except
when one body is composed of wood and the other of metal, in
which case the resistance increases with the velocity.
E. S. 20
154 FKICTION.
(6) It is also found that friction is diminished: by oiling
and polishing the surfaces in contact. There is a limit however
to the latter, for if they be very highly polished, the resistance
increases.
(7) The friction of cylinders rolling on planes, is propor-
tional to their pressures directly and their radii inversely.
It is remarkable, that friction of this kind, unlike that
between two planes, is not diminished by greasing or oiling the
surface of the planes and cylinder. This kind of friction is much
less than that produced by rubbing.
CHAPTEE IX,
ON ELASTIC STEINGS.
237. Strings made of certain substances are found to be
elastic ; that is, they admit of being lengthened by the^ appli-
cation of forces to their extremities, and regain their original,
dimensions, or nearly so, when the forces are removed. Spiral
springs composed of steel wire, such as the one exhibited in
fig. 71, are found to possess the same property in a remarkable
degree. The connection between the force which stretckes a
string, or a spring of the kind here mentioned, and the increase
of length cannot be investigated from mathematical considera-
tions, but is to be determined entirely by experiments.
Let MN (fig. 72) be a very smooth horizontal table ; AB
an elastic string or spring laid upon it and fastened at A ;
W a weight stretching the string by means of a thread passing
over the pulley G, whose position is such that ABG coincides
with the table. Then, if W stretches the string to h, and another
weight W stretches it still farther to V, it is found that
Bh : BV :: W : W;
that is, the excess of a given elastic string or spiral spring above
its natural length is proportional to the weight which stretches it.
238, This excess is, in different springs of the same make and
materials, proportional' to their lengths.
For the tension of a string being the same in every part,
if we divide the string into any number of equal parts, the
increase of length in each part will be the same, and therefore
the increase of the whole string will be proportional to the
number of these equal parts which it contains : that is, to its
length.
156 ELASTIC STRINGS.
239. Consequently, upon the whole, the increase of length
of a string is proportional to
(its length) x (weight which stretches it).
Wherefore if L lie the natural length of a string, and I its
length when stretched by a weight W,
l-LcxL.W=C.LW;
where G denotes a constant dependent on the material, thickness
and make of the string.
240. Suppose the string AB (fig. 73), whose ncctural length
is &, to he suspended vertically from one end A, and stretched by
its own weight w only ; to determine the increase of its length.
In AB take any points F, Q very near to each other, and
when the string is stretched let b, p, q, a be the points cor-
responding to £, F, Q, A; x = BF, Sx = FQ, y=bp,By =pq.
Then Sx is stretched into S^ by the weight of bp or BF
■,. 1 wx
which = — ;
a
.'. By — Bx= C.Bx
" wx
J
a
therefore, dividing by Sx, and taking the limits.
wx
d^-l=0.^;
a
.'. y — x = — . — , by integration ;
.', ab — AB = — . = 4 Cwa.
2 a 2
Hence the increase is one half of what it would be, ifKQ were
stretched upon a horisontal table by i weight equal to its own
weight.
2.41. A weight W is now suspended Jrom, b, to determine the
further increase of length.
The weight which stretches jjg" is, in that case,
a
ELASTIC STRINGS. 157
.-. al-AB= CWa + ^Cwa.
242. Of this increase the part ^Cwa we have seen is
due to the weight of the string, and therefore GWa, the part
due to the weight W, is the same as if the string had no
weight. Hence when a string is stretched by several forces,
each one produces the same increase of length as it would do if
the other forces did not act.
By way of illustration we shall add the following ex-
amples.
243, Two weights P, Q (fig. 74) resting .on two incUned
planes AB, AC, are connected hy a given elastic string; to find
the position of equilibrium.
Let a, yS be the inclinations of AB, A G, and 6 that of PQ
to the horizon; a=the natural length oiPQ; r= its tension.
Then P is kept in equilibrium on the plane AB by the force T
acting in the direction PQ ;
.-. T cos APQ = P sin a, (Art. 212) .
But^P^=a-^;
„_ P sin g
cos (a — ^) '
a- -1 1 m Qsin/3
Similarly, T^ ^^^^^y ,
p cos (j8 + g) _ ^ cos (g-g) .
cos g sin /8 cos sin g '
.-. P (cot ^-i&ne)=Q (cot a + tan 6) ;
.-. tan = p , ^ , which gives ;
SiniPQ = a+C.a.T
=''1 +cos(«-g)r
158 ELASTIC STRINGS.
From which PQ is known and thence AP and AQhj means
of the triangle APQ, whose angles are all known.
244. Two egual weights P, Q (fig. 75) are connected hy an
elastic string, whose natural length is BC ; to find the nature of
the curves BP, CQ, on which they will always rest in equilibrium
with the string parallel to the horizon/ the plane of the curves
being vertical.
It is manifest, since the weights are equal, that the curves
must also be equal. Bisect BO in A, and draw AM vertical ;
AB = AO = a, AM= x, MP=:MQ= y, T= the tension of P^ ;
.-. PQ-BG=G.BG.T,
or 2y-2a== C'la. T;
.'. y — a= CaT.
But P being sustained upon the curve BP by its gravity P
and -the force T, we have by Art. 213,
T=Pd^;
.'. y — a= GaPd^ ;
.*. {y — a)' = 2 GaPx, by integration,
which is the equation of a parabola. Hence BP, GQ are two
semi-parabolas, whose vertices are B, G.
CHAPTEE X.
ON THE FUNICULAR POLYGON, ON THE OATENAEY,
ON ROOFS AND BRIDGES.
ON THE FUNICULAR POLYGON.
245. ABCDEF (fig. 76) is a cord, supposed devoid of
weight, suspended, from two points A., Y in a horizontal line;
at the knots B, C, D, E weights, W,, W,, W,, W^......are
hung; to determine the proportions of these weights that it may
hang in a given form.
From A draw Ac, Ad, Ae, Af respectively parallel to the
portions BG, CD, BE, EF of the cord ; and denote the re-
spective inclinations of jdJ?, 5 C, GB to the horizontal line
AF hj a, )3, 7 ; draw MB vertical. Then B is kept at
rest hj the tensions of AB, BG and the weight W^, which
forces are respectively parallel to the sides BA, Ac, cB of the
triangle ABc, and are therefore proportional to them. Therefore
W^ is proportional to Be. In the same manner W^ is proportional
to cd; and they are on the same scale, for in both Ac represents
the tension of BG.
W^_Bc^ BM-cM
''■ W~ cd" cM-dM
AMi&Ta. a - AMixa ^
~ ^ilf tan p — ^ilf tan 7
tan a — tan ^
~tan)8 — tan 7'
„. ., 1 PFj tan^-tan7
Similarly, ^ = ^-— ^-^
160 THE FUNICULAR POLYGON.
It appears, therefore, that any one of the weights is propor-
tional to the difference of the tangents of the angles at which the
two sides of the polygon, which form the angle at which it is
suspended, are inclined to the horizon.
The angles MAe, Maf, which are situated above the line AF,
are to be accounted negative.
246. The horizontal tension of any part of the string is repre-
sented by AM, for it is the resolved part of the lines AB, Ac,
Ad which represent the whole tensions ; and this horizontal
tension : any weight ( W^ suppose)
:: AM : cd :: \ : tan ^8 — tan 7.
Cob. The tension of any string BG : the horizontal tension
:: Ac : AM :: ^If sec /3 : AM :: sec /8 : 1.
247. liAB,BC, GD in the preceding figure, instead
of being lines devoid of weight, be heavy beams of wood, or bars
of metal, connected at the joints A, B, G, D by hinges, we
must consider each beam as exerting by means of its weight
vertical forces at its extremities. Thus, if Wj, w^, w^ be the
weights of AB, BG, OD we may consider BG as exerting
equal pressures ^w^at B and C in a vertical direction, the centre
of gravity of the beam being supposed at its middle point ; in
like manner AB exerts a vertical pressure equal to Jwj at B, and
therefore we may consider W^ + ^{w^+w^) as the whole weight
suspended at B. Similarly, the weights to be considered as
suspended at G, D are respectively
and these weights are to be used instea'd Of those given in the
preceding articles.
These considerations are intimately connected with the con-
struction of suspension bridges.
248. If TFj, W^, Tfg are evanescent, then the weights
to be considered as suspended are ^(Wi + Wj), \{w^ + w^
and if the beams are all equal, each of these become equal to w^.
ROOFS AND BRIDGES. 161
ON ROOFS AND BRIDGES.
249. If the whole figure of Art. 245, be inverted or turned
round the horizontal line ^i^ through an angle of 180°, as in fig.
77, we shall find the same relations between the weights as be-
fore ; jt will also appear, from the same reasoning as in Art, 247,
that the weights to be considered as hanging from B, G, D
are the same as there investigated. In this state the problem
contains the whole theory of roofs, arches, and bridges. If
ABGBEFhe considered as a roof, of which AB, BG. are the
beams, then the horizontal thrust at A and F tending to push
out the walls on which the roof is erected, is represented by AM,
on the same scale as that wherein Be represents the weight to
be suspended from B; it is therefore equal to
TF. + IK + w,) ^
tan a — tan yS
This thrust is usually prevented from taking effect upon the
walls by inserting the ends. A, F of the beams AB, FE into
another AF called the tie-leam, which is thus made to sustain
the whole thrust ; at other times the walls are prevented from
bulging by buttresses, or shores, built against them.
If it were required to construct a roof of given span with
given beams, which has to support given weights, we must take
an equal number of smaller proportional beams, and connect
them by strings or pins at the joints, so as to allow them "to
move freely, and load them with proportional weights. Then
if this model be suspended from its extremities at a propor-
tional distance, as in Art. 245, it will assume the required
form, which we have merely to turn round AF through an
angle of 180°, and it will be a perfect model of the required
roof; and will possess the property of being in equilibrium
in every part. In such a roof there will be no unnecessary strain
on any part of the materials of which it is constructed, and con-
sequently no part will require to be unnecessarily strong. In
this simple manner we may also obtain the model of a bridge of
given span, by taking a great number of very short beams to
represent the arch stones, and connecting them as before. If
E. s. 21
162 THE CATENARY.
when we suspend this model-string of arch stones loaded with
weights proportional to what (in the place they occupy in the
bridge) they will have to sustain, we find that the bridge would
be too lofty, we niust remoye the points of suspension farther
apart, until we have obtained the proper altitude. This method-
will give us a bridge, in perfect equilibrium in every part, and in
which there is, therefore, no injurious strain, no useless strength,
nor dangerous weakness in any part.
ON THE CATENARY.
A Catenary is the curve assumed by a fine chain or flexible
string when suspended from its extremities.
250. To investigate the equation of the catenary.
Let AOF (fig. 78) be the catenary; A, F being the points
from which the chain is suspended, and being the lowest
point of it. Through draw BOD vertical, which take for
the axis of x, being the origin. From P any point of the
chain draw PM perpendicular to OD ; and draw PT a tangent
at P. x= OM, y = MP, s = OP. Since there is equilibrium
we may suppose the part OP to become rigid ; then since it is
kept in equilibrium by the action of three forces (its own weight
and the tensions at Pand 0), which act upon it in the directions
of the sides of the triangle MTP taken in order, we have
tension at PM T>m\r j
weight of OP MT "^
But if the chain be uniform, the weight of OPmaj be repre-
sented by its length s, and the tension at by the length of a
piece of the same chain of the length a ;
.-. 1
'Jd+r
x+C='U+?.
THE CATENARY. 163
But when x = 0, s = 0, and therefore 0=a;
.-. x + a = '/^T7,
and .'. 03° + 2ax = s",
which is the relation between any arc and its abscissa.
251. To find the equation of the catenary in terms of the
rectangular co-ordinates x, j.
The equation required is expressed in its most simple form
hj taking for the origin of co-ordinates the point £, which is
such that OB=a. Let then BM=x; then from the last
article,
\la?—a^ = i
■d^s
IX —a=s;
X
.-. y = a log, {x + 'Ja?-a^) + C.
But when x — a, y = 0, consequently 0= — a log, a ;
, x + 's/a^-a'
•'• 3/ = «log. •
This is the equation required.
252. The relation between x and y, and that between s and
y may be expressed in very simple exponential forms as follows.
From the last equation we have
a \a I
a \a J
164 THE CATENARY.
-=6" +6- (1.
a
Again, - = 2^-,-.
:e — e
(2).
253. Def. If through 5 we draw BG horizontal, it is called
the directrix oi the catenary.
254. The tension of the chain at any point P is eqyt/xl to the
weight of apiece of the same chain of the length BM.
J, tension at P 2!P 1
tension at PM sin PTM
yi + id^yf
a
X
a
_ weight of lengtii x
~ weighf of length a '
But the denominators of these fractions are equal ;
.•. tension at P= weight of chain of length x.
255. We have supposed the chain to be uniform; if it
should be of variable density or of variable thickness, let p be
such a quantity that phs may represent the mass of a small
element (S« =) PQ of the chain. Then the weight of OP is
S {gpZs) =gjpds = gjypd^; and representing the tension at by
ga, we have by proceeding as in Art. 250,
- tension at
weight of OP
ga
THE CATENARY. 165
.'. pd^ = adyX (1).
When jO is given, this equation being integrated will give the
form of the catenary.
256. To find the law of density that the catenary may he of a
given form.
In this case the relation between x and y is given to find p.
From the last Article we have
Now the quantity which is multiplied into p is the radius of
curvature of the curve at P, and d^ is the secant of the inclina-
tion of the tangent at P to the horizon ; wherefore
_ a . sec^ of the inclination
'^ radius of curvature
257. To find the form of the catenary when the chain is acted
on hy a force tending to a fixed centre.
Let BAG (fig. 79) be the catenary, suspended fi"om B, 0.
8 the centre of force, A that point of the chain which is nearest
to 8; therefore 8A is a* normal at A. Let P be any point,
and PQ a small element of the curve, a = 8A, s = AP, Ss = PQ,
r = 8P, r- + Sr = 8Q, t = tension of the chain at P, t + Bt= the
tension at Q, pBs = the mass of PQ, and F= the force which
acts at P towards 8. Then the weight of PQ = FpBs, which
we may suppose to act ultimately in the direction Q8. Hence
resolving the forces, which act upon PQ, in the direction of the.
tangent at Q or P, we have
t + FpSs cos PQ8='t + Bt.-
But if we draw PP' perpendicular to 8Q, we have
Bs cos PQ8 = Br ;
.-. FpBr = Bt;
.:JFpdr = t (1).
166 THE CATENAEY.
The integral is to be taken from r= 8A to r= 8P.
Again, the chain AP is kept in equilibrium by the tensions
at A and P, and by the weight of each particle of it in a direc-
tion passing through 8. Hence taking the moments of all these
forces about 8, we have
a . (tension at A) =pt,
where p is the perpendicular from 8 upon the tangent at P.
The left hand member of this equation is constant, and there-
fore representing it by C, we have
«=- (2).
P
JFpdr = ^;
Hence, combining equations (1) and (2), we have
i'
,.Fp = -^.
P
When jO is given, this equation being integrated will give the
form of the catenary.
258, To find the law of density that the chain may hang in a
given form when acted on hy a given central force.
In this case the relations between F, r and p are given to
find p. From the last Article we have
Cd,p
259/ Cor. Since QF = hr, FpSr = the weight of a piece
of the given chain of the length QP' and density p ; hence if the
density of the chain be the same throughout, the equation (1)
taken between its proper limits gives
tension at P— tension at A
= weight of chain of the length P8
— weight of chain of the length SA.
This result corresponds to that obtained in Art-. 254.
THE CATENARY. 167
260. To find the form of the catenary when the chain is acted
on by any forces in its own plane.
Let AB (fig. 80) be the curve, in the plane of which take
two lines Ox, Oy perpendicular to each other for co-ordinate
axes. Let X, Y be the components of the accelerating force
which acts at P, parallel to Ox, Oy respectively. . Let FQ be a
very small arc; FT, QT tangents at P and Q meeting in T.
From T draw TG a normal to the curve.
Let X = OM, y = MP; s = AP, Ss = PQ; x + Bx, y+ Sy
the co-ordinates of Q; and pSs the mass of the arc PQ. We
suppose Bs so small that the accelerating forces X, Y may be
considered the same for every point of it ; consequently XpSs,
YpSs are the weights of PQ estimated parallel to Ox, Oy re-
spectively. Now PQ is kept at rest by three forces, the tension
t at P, the tension t + St at Q, and the resultant of XpSs, . YpBs ;
consequently, as PQ may be considered rigid without disturbing
the equilibrium, these three forces all pass through the point T;
they therefore satisfy the conditions of equilibrium of forces
acting on a point. Eesolve the forces parallel and perpendicular
to the normal GT;
.: = < cos PTG + {t + Bt) cos QTG- XpSs . d,y + YpBs . dfc,
and
= « sin PTG -{t + Bt) sin QTG- XpBs . dpe - YpBs . d.y.
Now cos PTG = -^^— = hBs 'iidfW+WW '
rad. curv. ^ ^ ' '^ "" '
and sin PTG = 1 ultimately ;
hence by substitution and dividing by Ss, we have
= t '/{dWWW - ^p^.y + '^9^?'^
and = d}, -\-Xpdfja + Ypd,y.
By eliminating t between these equations, we obtain the dif-
ferential equation of the required curve.
[The remaining Articles of this Chapter are from the pen of
the Eev. J, A. Coombe, Fellow of St John's College : by whose
permission they are here inserted.]
168 THE CATENARY.
261. Peop. To find the fiyrm of eqiiilibrmm of a unifirm
tnextensible string on a surface and acted on hy any forces.
Let M = be the equation to the surface, xyz the rectangular
co-ordinates of any point in the string, and therefore of a point
in the surface; s the length of a portion of the string inter-
cepted between a fixed point in the string and the point {xyz) ;
XYZih& resolved parts of the forces at the point {xyz) parallel
to X, y, z: R the normal reaction at the point {xyz), making
angles aySy with the axis of co-ordinates ; T the tension of the
string at the point {xyz), one extremity of an element Zs of the
string, and acting in the tangent at that point.
Hence Td^ will be the resolved part in x, and
Tdpi+d,{Tdp;).Zs
will ultimately be the resolved part in x of the tension at the
other extremity.
Hence d, {Tdp;) Ss will be the difference of the resolved parts
in X.
The other forces acting on Ss parallel to x are XSs and
liSs cos a, supposing the forces to act equally on every point of
the very small element Ss.
Now Ss being at rest under the action of these forces, we may
suppose it to become rigid and apply the equations of equilibrium.
Hence we have (dividing by Ss),
d,{Tdp:) + X+Rcosa=0 (1).
Similarly, d,{Td^)+ F+^cos/3 = (2),
and d,{Td,z) +Z + Bcosy = (3).
The equations of moments may be dispensed with for the
following reasons. Consider three adjacent points P, Q, B, in
the curve, Q being in the middle, and the tangents FT, BT,
meeting in T. Then the plane containing these points and the
tangents PT, BTvnR be the osculating plane.
Now the forces X and B cos a being supposed to act equally
on every point of PB will have a resultant through Q, and so
will the like forces parallel to y and z.
THE CATENARY. 169
Hence the whole of the forces acting on PR being reducible
to three, lying in the osculating plane, will pass through a single
point T in that plane; and the equations of moments taken
about this point will be identical.
Recurring to the above equations, our object will be to elimi-
nate T and R between them.
Now (1). d,x+ (2). d,y + (3). d,z = gives
d,T-\- Xd^x + Yd,y + Zd^ + R (cos adjc + cos ^d.y
+ cos^c?/) = 0,
since {dp-f + {d,yf + {A/i' = l;
and .■ . dfis. d^x + d,y. d^y + d,z . d^z = 0.
Also because the tangent to the curve is perpendicular to the
normal to the surface, we have
cos a . dp: + cos /3 . d,y + cos 7 . <Z,a = 0.
Hence the above equation becomes
d,T+ Xdp: + Yd,y + Zd^ = 0;
or if Xd,x + Yd,y + Zd^ = dp,
wehave ir+v=0 (4);
(the arbitrary constant being included in v).
Again, (1). d,y - (2). dps = gives
T(d,ydjx - dpd^y) + Xd,y - Yd,x + R {cos a.d,y
— cos ^djx} = 0.
Now if i = {{d^uY + {d,uy + {d,uY]K
wehave con a.-Vd„u, cos /3 = FcZ^m, cos 'i = Vd,u,
the differential coefficients of u being partial.
Hence the above equation becomes
T {d.yd1x - dpdfy) + Xd.y - Yd,x + RV{d,ud.y
-dyud,x)=0 (5).
E. s. 22
170 THE CATENARY.
Similarly,
T [d^d^z - d,zd^x) + Zd^-Xd^ + R V{d,udp:
-d,ud,z)=Q (6),
and
T {d^dfy - d.ydfz) + Ydp - Zd.y -^RV {d^ud^
-d,ud.y)=0 (7).
Hence (5). d,u + (6). d^u + (7). djii = gives on substituting
for Tits value derived from (4),
vd,u{d,yd!x-dp:d^y)\ ( d,u {Yd,x - Xd.yh
+ vdyU {dpd^z — d^d^x)\ = ■!+ dyU {Xd^ — Zd.xy, [A).
+ vdji {d^d^y — d,yd^z) J 1+ dji {Zd,y — Yd^) )
This equation together with m = are the equations to the
curve of double curvature into which the string is arranged.
262. CoE. 1. Suppose the resultant K of the forces XYZ,
acting at the point {xyz), is in the normal to the surface at that
point, so that
X=KVd,u, Y=KYdyU, Z=KVd,u,
the equation (-4) then takes the form
d,u {d,yd^x — dp:d^y)\
+ dyU {d^d^z-d^d^x)\ = (i (8),
+ d^u {d,zd^y - d.yd^z) J
or substituting A, B, G for the coefficients of d^u^ dyU, d,u, the
equation is
Ad^u + BdyU + Cd,u = 0.
Now the equation to the osculating plane at the point {xyz) is
A{d -x) + B {y' -y) + G {z -z) = Q (9),
and the equations to the normal are •
4m dyU ^ d,u .^Q,
x' — x y' — y z' — z ''
x'y'z being the current co-ordinates of a point in the plane or
normal: and when the plane (9) contains the normal (10) we
have the condition
Adjt + BdyU + Cd^u = 0.
THE CATENARY. 171
Hence equation (8) expresses that the osculating plane con-
tains the normal: now this is the property of the shortest line
between two points on the surface.
263. Prop. To find the pressure on the surface at any
point.
(1). Cos a + (2). cos /3 + (3) . cos 7 = 0,
gives —B = Xcos a. + Fcos /3 + Zcos 7 + T{d^x cos a.
+ d^y . cos /3 + d^z cos 7}.
Now if p be the radius of absolute curvature at the point
{ocyz), and X, fi, v the angles it makes with the axes, we have
cos X = pd^x, cos /i = pd^y, cos v = pd^z ;
.*. — BSs = XSs cos a + YSs cos /8 + ZSs cos 7
+ T. — (cos a cos \ + cos /8 eos /* + cos 7 cos v).
r
Let 9 be the angle between the radius of absolute curvature,
in the direction of which the resultant of the tensions on the
extremities of Ss acts, and the normal to the surface, then
cos = cos a cos X + cos /S cos p, + cos 7 cos v.
Substituting in the above equation, we have then
pressure on a portion Bs of the surface
= resolved force in the normal
+ resolved tension in the normal.
264. COK. When there are no forces acting on the string,
so that X = 0, r= 0, Z= 0, we have
d,T= 0,
' or T= constant = k,
and pressure = - on an unit of length,
1
172 THE CATENARY.
265. Pkop. To find thefirrm qfequilihrmm when the sVring
is not attached to a surface.
The equations (1), (2), (3), will give the equations of equili-
brium, by putting B=Q : and eliminating T between (1), (2), (4),
and also between (1), (3), (4), we have the two following equa-
tions to the form of the string.
V {dpd^y - d,yd^x) = Ydpi - Xd,y\
V {dpcd^z —d^d^x) = Zdp — Xd^)
(.B).
266. Cor. In the case of gravity, supposing the axis of a
vertical and measured upwards, we have
X=0, F=0, Z=-g;
.: dficd^y-d,yd^x = Q.
This is the differential equation to a straight line in the
plane xy, so that the chain hangs in a vertical plane. Take this
plane for the plane of xz, and the lowest point as origin. We
have, since
d,T=gd^, and .". T=g{z+c),
{z + c) {dfad^z — d^zd^x) = d,x,
or since d,xd^x + d^d^z = 0,
d^ d,^x_
;■■■ ^™ ~"'
z + c djx
or z + G= cd^s ; since when z = 0, d,x = 1.
Hence d^ =
^l{z+cf-,^'
.-. x+C=c log, {z+c + ^{z + cf - c"},
and when s = 0, x = 0; .: C=c\og,c; and
6« =i±-£.
c
THE CATENARY. 173
c / - — \
267. Cor. 2. The Catenary also possesses the property of
being the curve of total minimum tension, supposing gravity
alone to act.
Thus tension =5^ (s + c).
Hence« to find the curve having the above property, we
have
/^ {z+c)d^ = z, minimum,
or when /j, (s + c) . Vl + {d^Y = a minimum.
This is the case when
by the principles of the Calculus of Variations; or when
z + c = cd^, and this has been just shewn to be the differential
equation to the Catenary.
CHAPTER XL
PROBLEMS.
1. Given the magnitudes of two forces which act on a point,
and the angle between the lines in which they act; to find the inag-
nitude of their resultant.
Let P, Q be the two forces acting upon the point (fig. 81)
in the directions OA, OB. Take OA, OB to represent them, and
complete the parallelogram OBCA ; the diagonal represents their
resultant R.
Let 0. = A OB the given angle.- Then from the triangle
OA G we have
0C^= 0A''-20A. AC cos OAC + AC
=^0A' + 20A. OB.cosPOQ + OB';
.: R' = P' + 2PQcos(i+ Q".
2. Three forces acting on a point are found to balance each
other when their directions make angles 105°, 120°, 135° with each
other. Find the relation of the forces to each other.
Let F^, F^, jPj be the forces respectively opposite to the
angles 105°, 120°, 135°. Then by Art. 28, we have
F^ : F^ : F^ :: sin 105° : sin 120° : sin 135°
:: cos 15° : cos 30° : cos 45°
:: cos (45° - 30°) : cos 30° : cos 45°
V3 + 1 VS 1
2 V2 ■ 2 ■ ^2
:^ *
PROBLEMS. 175
F F F
V3 + 1 V6 2 '
for which when the magnitude of one of the forces is given, the
magnitudes of the other two are known.
3. A weight W is sustained upon a smooth inclined plane
hy three forces each equal to JW ; one acting vertically upwards,
another parallel to the plane, and the third. in a horizontal
direction; required the inclination of the plane to the horizon.
(Fig. 82.)
Let C be the body placed on the inclined plane AB]
F^, F^, jFg the forces mentioned in the question. Besides
these, G is acted on by gravity which is equal to W and acts
in the downwards direction CW, and by the re-action R of the
plane, which because the plane is smooth, acts in a direction
GR perpendicular to AB. Hence the body G is kept at rest
by five forces, all of which act in the same plane; hence the
conditions of equilibrium are (Art. 32), that the sums of the
resolved parts of these forces parallel to two lines in the plane
of the forces shall be separately equal to zero. Eesolve them
in the directions of GB, GR ;
.: = RcosRGB+F,cosF^GB + F^cQsF^GB+F^cosF^GB
+ Wcos WGB,
and
= Rcos RGR + F, cos F,GR + F, cos F,GR + F,cosF,GR
+ Wcos WGB.
If we put 6 for BGF^ the required inclination of the plane,
these equations become /
= i^isin^-|-j?; + i^3cos^- Wsind,
and O = R + F^cos0-F^sm0- Wcos9 (A).
The former of these, observing that F^ = F^ = F^ = ^W,
gives
1 +cos5 = 2sin0.
176
PROBLEMS.
or 2 cos'' - = 4 sm - cos - ;
tan 2 =
i;
.-. e =
2 tan"'
(i)
53°. 7'
. 48".
Remark. The reader will observe that we have ohtained
this result without making use of the second condition (A)
of equilibrium. From that equation we might have deter-
mined R, the pressure of the body upon the plane ; but as that
is not required by the enunciation of the problem, we make
the important remark, that it is not always necessary to employ
all the equations of equilibrium in solving a problem : and in
resolving forces, our aim must be to resolve them in a direction
at right angles to such forces as are not known and not required
to be found.
The directions in which we may resolve the forces are quite
arbitrary (Art. 98) ; we might, for instance, have resolved the
forces parallel and perpendicular to the horizon, from which
would have resulted the two equations
O = i?sin(9-i^,cos5-i?;,
and = Rco&e + F^ + F^s\nd-W;
but here we see the unknown force R is involved in both the
equations of equilibrium; and in order to solve the proposed
problem it is necessary to eliminate R between them : this
necessity is avoided, and the result at once obtained, by resolving
the forces in a direction at right angles to that in which R acts.
We shall generally avail ourselves of this artifice in the problems
which follow.
4. Three equal forces act upon a point in the directions
of three lines which include angles 105°, 120°, 135°; fnd the
magnitude and position of their resultant.
Since the sum of the given angles = 360°, the forces all act
in one plane. ~Let F^F^F^ (each equal to P) be the three forces
PBOBLEMS. 177
acting on the point (fig. 83), the angles F^OF^, F^OF, being
120°, 135° respectively. Produce F^O to x, and in the plane of
the forces di-aw Oy perpendicular to Ox, Then if B be the
resultant, and 6 the angle which its direction makes with Ox,
we have, proceeding as in Art. 81,
. Iicos9 = -F^ + F^cosF^Ox + F,cosF,Ox
= - P+P cos 60° + Pcos 45°
_ V2-1
~^' 2 '
i? sin ^ = i?; sin 60° - F, sin 45°
_p V3-V2
~ 2 '
.-. tan = ^^^=^ = •7673269;
.-. ^ = 37°. 30',
which determines the position of B ; and the equation
^ = {RcoseY+{RsmeY
= P'x -0681484,
or i2 = Px -2610525,
gives the magnitude of B.
5. If three forces proportional to the sides of a triangle he
applied perpendicularly at their middle points, they will balance,
supposing them all to act in the plane of the triangle.
Let ABC (fi^. 84) be the triangle ; a,h, c the middle points
of its sides. At these points apply three forces F^, F^, F^ re-
spectively proportional to the sides on which they act, in direc-
tions perpendicular to those sides. Then, because the sides of
the triangle are bisected perpendicularly, the lines F^a, Ffi, F^c
being produced will meet in the centre of the circle circum-
E. s. 23
178
PROBLEMS.
scribing the triangle. We may therefore suppose them to act
at : and because
F,:F^: F, :: BC : AG : AB
:: sin^ : sin5 : sin C
:: sin^.Oi^, : smF^OF^ : sin i?;Oi^„ .
therefore, by the converse of Art. 28, the forces balance each
other.
6. Two given equal unifiyrm heams AC, BC (fig. 85) having
their lower ends connected hy a string are placed in a vertical
plane, upon a smooth floor, their upper ends leaning against each
other. Required the tension of the string AB.
Let a be the length of each beam, i the length of the string
AB; 0=iCAB; G the centre of gravity of AC, which since
the beam is uniform will be in its middle point. The beam A G
is kept at rest by B the pressure oi BC against it in a horizontal
direction ; T the tension of the string AB in the direction AB ;
R' the reaction of the floor, which since the floor is smooth will
act at right angles to AB; and by W the weight of the beam
acting at G in the direction G W vertically downwards. (The
conditions of equilibrium for this case are stated in Art. 67.)
To avoid the force E, resolve the forces horizontally, and
take the moments about A ; then
= ^-2",
and (i=:R.AC&ine-W.AG(iQ&e;
.-. T=R = W^.cate
AG
W h
2 Wia'-h^'
7. A string PCP (fig. 86) having two equal weights P, P
fastened to its extremities, passes over a pulley C, and two pegs
A, B. A smooth heavy ring Q is passed over C : required the
position in which it will rest, its inner diameter heing such as to
PROBLEMS. 179
Tceep the parts of the string above it parallel; and the pegs A, B
being similarly situated with respect to C.
Let W represent the weight of the ring ; the inclination
ot AQ to the horizon. Because the ring is smooth the tension
of the string will be equal to F in every part; and when the
equilibrium is established we may suppose the ring and string
to cohere at the points of contact, by which supposition we per-
ceive that Q is pulled upwards by the two parts of the string
between the ring and the pulley ; and obliquely by the portions
of the string between the ring and the pegs. Hence the ring
is kept at rest by five forces, viz. the two vertical forces P, P
acting along the lines QO, Q'C; the two oblique forces P, P
acting along the lines QA, Q'B; and its own weight W acting
vertically downwards : consequently resolving them in a vertical
direction, we have
= 2P-2Psme-W;
•• sm0=l-^;
which determines the position of Q.
8. Two weights P, Q (fig. 87) are connected hy a string,
which parses over two smooth pegs A, B situated in a horizontal
line, and supports a weight W which hangs from a smooth ring'
through which the string passes. Find the position of equiU-
hrium : and also whether the equilibrium is stable or unstable.
Since the ring O is smooth, the tension of the string is the
same throughout, and therefore
P=Q.
Also we may consider the point C as kept at rest by three
forces ; the tensions of CA, CB, and the weight W; hence by
Art. 28,
P F _ W
8mB0W~ sin AGW~ sin A CB '
.:AGW=BCW;
180
and
PROBLEMS.
W ainACB
F sin AGW
sin 6
~ . e
sin-
= 2cos-,
d representing the angle AGB. This equation gives 0, from
which the place of the ring is known.
Again, let 2a = AB, 25 = the length of the string, c= CW,
i the distance of the common centre of gravity of P, Q, TT be-
low AB; then producing WC to meet AB in 1), we have
(P+ Q + W)s=F.AP+Q.BQ + W.DW;
.-. (2P+ W)z = 2P./b 1-
_^\+TF.(c + acot|);
ff -I ff
.: (2P+ W) d^ = Pa cosec- cot - - - Wa cosec^-
= I cosec'' I (2Pcos | - PT) = ; (Art. 168) ;
a a Z
.: (2P+Tr)£?e=i = -^cosec''|.Psin|,when TF=2Pcosf.
Hence, in the position of equilibrium i is a maximum, and
therefore the altitude of the common centre of gravity is a
minimum, consequently (Art. 169) the position is one of stable
equilibrium.
a
Remark. The equation d^ = 0, gives Tr=2Pcos- for the
condition of equilibrium, and consequently the latter part of the
preceding investigation includes the solution of the whole ques-
tion proposed ; the first part therefore might have been omitted ;
but we have inserted it as an example of the application of
Art. 28.
PROBLEMS. 181
9. A straight rod rests with its ends upon two given smooth
inclined planes, in such a position that the vertical plane which
passes through the rod is at right angles to the given planes; find
the position of equilibrium of the rod.
Let AB (fig. 88) be the rod, Q its centre of gravity; a, y8, 9
the inclinations of the planes OA, OB, and of the rod to the
horizon. R, B' the reactions of the planes at A, B, which will
he in normal directions because the planes are smooth. AG = a,
BG = b. The forces which keep the rod in equilibrium are the
reactions B, B' and its own weight. To avoid the weight of the
rod, resolve the forces parallel to the horizon; and take their
moments about G ;
.•. = ^sina — ^'sin/3,
and 0=^B.GA sin GAB - B' . GB sin GBB'
= Ba cos OAB - B'b cos OBA
= Bacos {a. + 0) -B'bcos (/3-0) ;
a cos {a+ 6) _^ b cos {^—ff)
sin a sin /3 '
, J. /I a cot a — b cot B
and .". tana = — — , ■^ .
a+b
Remark, In the enunciation of the preceding problem it is
assumed, that the vertical plane which passes through the rod is
at right angles to the given planes. This is a particular case of
a more general proposition which we shall now give.
10, ^ a body of any form whatever rest in equilibrium upon
two smooth inclined planes, the line of intersection of the planes
must be horizontal.
For distinctness' sake let the planes be called A and B.
Then as they are smooth the reactions of the plane A, at the
various points of contact of the body with it, are all perpendi-
cular to the plane and therefore parallel to one another; and
consequently their resultant {B suppose) is (Art. 71) also per-
pendicular to A.
182
PROBLEMS.
Similarly {B') the resultant of the reactions of the plane B
upon the body is perpendicular to B. Consequently we may
consider the body as kept at rest by the action of three forces ;
viz. B and B' (acting respectively perpendicular to the planes A
and B) and the weight ( W) of the body acting vertically at its
centre of gravity. Consequently, by Art. 96, the lines in which
these forces act lie in one plane ; and as the vertical through the
centre of gravity is in this plane, therefore the plane is a vertical
plane ; and as normals to the planes A, B, in which B and B'
act also lie in it, therefore it is perpendicular to each of them,
and consequently also to their intersection : that is, the line of
intersection of A and B is perpendicular to a vertical plane, and
consequently it is horizontal.
CoE. If a body of any form rest in equilibrium upon two
given inclined planes with one point only in contact with each
plane, the vertical plane which passes through the two points
of contact will be at right angles to the given planes, and pass
through the centre of gravity of the body.
11. LM is a smooth sphere of radius r (6 inches) and weight
w (3|^ Ihs.), in contact with a plane AM inclined to the horizon at
an angle a (60°). AB is a beam of weight W (lOOlbs.), and
length a {6 feet), moveable about a hinge at A, and by its pressure
preventing the sphere from descending down the plane. Determine
the positions of the beam and sphere. (Fig. 89.)
Let B be the reaction at L between the sphere and beam ;
and B' that at Jf between the sphere and inclined plane; since
the sphere is smooth, the former acts in a direction perpendicular
to AB; and the latter in a direction perpendicular to AM. Let
'ie = ^BAM.
We may consider AB as a lever, whose fulcrum is A, kept
at rest by i2 at i in the direction GL ; and W at G, the centre
of gravity of AB, in a vertical direction ;
.-. B.AL= TF. 5 cos (a + 20),
or B. r cote =W.^cos (01 + 20).
PROBLEMS. 183
The sphere is kept in equilibrium by its own weight acting
downwards at G, and the reactions RR' in the directions LG,
MG. To avoid R' resolve these forces parallel to the plane ;
.•. = i? sin ^0—w sin a.
Hence, eliminating R,
2wr sin a cot 5 = Wa sin 20 cos (a + 2ff} ;
.'. sin" 6 cos (a + 20) = ™r- . sin a.
By substituting for W, w, a, r, a their values, we find from
this equation
0=4°, 45'. 30";
and .-. a + 20 = 69° . 49',
which is the inclination of the beam to the horizon.
The position of the ball is known from the equation
AM= r cot
= 5-822314 feet.
12. A uniform heavy rod CD rests with one end D on a
smooth inclined plane DB, and the other is suspended hy a string
of given length from a fiooed point A. Find the position ofequili-
hrium. (Fig, 90.)
Draw AB perpendicular to the plane ; and let ^, be the
angles which A C, GD respectively make with AB ; let a be the
inclination of the plane DB to the horizon ; let G be the centre
of gravity of the rod; a= GG = DO, b = AG, o = AB, R the
reaction of the plane at B, which since the plane is smooth will
be in a normal direction; 7= the tension of the string GA.
Since AB = the sum of the projections of A G, GD upon it ;
.". c = &cos^ + 2acos0 (l).
To avoid the weight of the rod, resolve the forces in a hori-
zontal direction, and take their moments about G ;
.'. a = Rama— T sin (^ — a),
and Q = RaBm0— Ta sin {0 — tj}) ;
184 PKOBLEMS.
.•. sin 6 sin ((^ - a) = sin a sin {& - ^) ;
.-. 2cot<^ = cot^ + cota (2).
But J cos ^ = c — 2a cos 6 from (1) ;
.•. 2b sin <j}={c — 2a cos 6) (cot 6 + cot a) ;
.-. iJ*" = (25 cos ^)' + (2J sin <f>Y
= (c - 2a cos 0)'' {4 + (cot + cot a)'}.
From this equation must be foimd by approximation, and'
then <]) will be known from (2).
13. Three rods AD, AE, BC are connected hy hinges at
A, B, C; AE is vertical and fixed ai E, and AD horizontal. At
D a given weight W is suspended. Find the strain upon the
hinges. (Fig. 91.)
Since the rod BG has hinges at both ends, it is incapable of
exerting any action except in direction of its length ; let T be
the pushing force which it exerts upon the hinge B in the direc-.
tion GB, and upon G in the direction BG. Let the strain upon
the hinge A be resolved into the forces X, F in the directions
BA,AG. Let
d^^iABG, a = BC, h=^BD.
Then the rod AD is kept at rest by X, Y, T and W; resolve
them vertically and horizontally, and take their moments about B;
.: 0=Tr-7'sina+F,
= Tcos a.-X,
and = Wi — Ya cos a.
From these three equations we find,
Wb
Y=
T=W.
a cos a
a cos a + b
a cos a sm a
a cos a + h
asm a
The first and last of these determine the magnitude and
direction of the strain upon the hinge A ; and the second equa-
PROBLEMS. 185
tion gives" the magnitude of the strain upon B or C; the direc-
tion of this strain has heen stated already to he CB for B, and
5(7 for C.
If the joints at A, B, were rigid, the action of BC not
heing necessarily in the direction of its length would be inde-
terminate: BG might even be removed without affecting the
equilibrium.
14. AB is a heavy uniform rod, moveable in a vertical plane
about a hinge A; a given weight P sustains the rod by means of
a string BCP passing over a smooth pin C situated in a vertical
through A and at a distance AC=AB. Fimd the position of
eguilibrium of the rod by the principle of virtual velocities.
(Fig. 92.)
Let W be the weight of the rod, a = AB its length, Q any
point in it ; draw QM perpendicular to AG. x=AQ, Q=i. BA C.
Then the virtual velocity of P
= d. GF= d {BGP- BG) = d (bGP- 2a sin |
6
= — d.2asm-= — acoa-.d6.
' ' The weight of a small element of the rod at Q, the length of
which is equal to Bx=W — ; and the virtual velocity of this
= d. GM=d{a — a; cos ^) = aj sin (
Hence for the whole rod the value o{'%{Fds) (Art. 113)
= t{W-.xsmede)
W ^
= --amed0.t{xBx)
W
= — sin dda Jxdx, from » = to x = a
= ^Wa sin dd0.
E. s. 24
186
PROBLEMS.
Hence, by the principle of virtual velocities,
. e p
Remark. The preceding solution is more difficult than is
absolutely necessary; we preferred giving it however as an
illustration of the meaning of the symbol % in Art. 113. The
following is more simple.
We may consider the weight of the beam as being collected
at its centre of gravity. Let Q be this point. Then by the
principle of virtual velocities,
(i = P.d.CP+W.d.OM
= P(- a cos^dd) + W.d{a-^ cos 6)
= - Pa cos 5 «?^ + PF^ sin Odd,
the same result as before.
15. Two heavy particles P, Q are connected by an injlexible
rod; and one of them (P) rests upon a given smooth inclined plane.
Required the nature of the curve on which the other (Q) must rest,
that there may he equilihrium in all positions.
Since there is equilibrium in all positions the common centre
of gravity of the bodies neither ascends nor descends (Art. 169),
it is therefore always at the same height above a given hori-
zontal plane. Let the equation of the given inclined plane be
y' = mx' (1),
the axis of x being horizontal, and that oiy vertical. Let x'y'
be the co-ordinates of P, and xy those of Q. Then denoting
the altitude of the common centre of gravity of P and Q above
the axis of x by b, and the length of the rod by a, we have
{P+Q)b = I),'+Qy (2),
a'^^ix-xy+iy-yT (3).
PROBLEMS. 187
From (1) and (2) we find
<3^z, Q.
which being substituted in (3), giye the following equation of
the required curve :
The values of the coeflScients of the first three terms shew that
the required curve is an ellipse.
16. A rod AB is placed in a smooth hemispherical howl, so
as to lean against the edge of the howl at P, with one end A within
it. Find the position of eguilihrium of the rod. (Fig. 93.)
Let C be the centre of the bowl, Q the centre of gravity
of the rod. The rod is kept in equiKbrium by the reaction
of the bowl at A, the direction of which passes through C;
by the reaction of the edge of the bowl at P, which will be
in a line PQ at right angles to AB; and by its own weight
acting vertically at G. There being but three forces, their
directions all paas through a point (Art. 96) ; hence QG \s, ver-
tical, and
AQ _ smAGQ
AG~BinAQG'
Let AG = a, and r = the radius of the bowl; then because
APQ is a right angle,. ^^ is a diameter of the sphere, and
therefore = 2r ; also if ^ = GPA, the inclination of the rod to
the horizon,
AGQ='rr-PGQ = 'rr-(^-e^='^ + e,
188 PROBLEMS.
and AQG = PGQ-PAG^(^-ej-e^'^-2e;
2r ^™U'"y cosg
a • /T „/i\ cos 25
sm
cos 5
2cos»5-l^
.-. cos''5--^cos5 = i:
from which equation is known.
17. ^ smooth heavy rod AB moveable in a vertical plane
about a hinge at A, leans against a heavy prop CD also moveable
in the same plane about a hinge at C. Find the position of equi-
librium. (Fig. 94.)
Let G, G' be the centses of gravity of l3ie two rods, at which
points we iaaj suppose their weights W, W to "be applied.
Let M' be the pressure which the rod AB exerts against the
prop; i? the reaction of the prop against AB; these forces
will be equal and opposite, and act in a line perpendicular
to AB.
The rod AB is a lever whose fidcrum is A, kept in equi-
librium by B and W; hence putting AG = a, GD = b, AG=c,
and the angles BA C, A GD =6, ^ respectively, we have by
taking the moments of B and W about A,
W.acosO^B.AD
= i2J.^ (1).
Similarly we perceive that the prop GD is a lever whose ful-
crum is G, kept in equilibrium by B' and W, hence taking
the moments of these forces about G, we have
PEOBLKMS. 189
W'.lcos<f>==B'bsinODB'
= m cos CD A
= - Bb cos (0+^).
Hence eliminating B by means of {1), we find
O = 2Wacos0 sin 6 cos {9 + (f>) + W'b cos ^ sin ^.
Also from the geometrical property of the figure,
c_sin(0+^)
b sinO
The last two equations will give the values of and <f>.
18. Two rods AB, AC rest against each other upon the
horizontal plane ED atK, and against two smooth vertical parallel
walls af B, C ; given the lengths and weights of the rods, to find
the distance of the walls when the angle between the rods is a right
angle. (Fig. 95.)
Let a, b be the lengths, and W, W be the weights of the
rods AB, A G, which we may suppose acting at their centres of
gravity. Let B, B' perpendicular to UB, DO he the reactions
of the vertical walls. = BAK Then the rod AB is kept in
equilibrium by W, B, and the forces which act upon it at A.
To avoid these last take the moments of all the forces which act
on AB aboui A ;
.-. O = W,^cos0-Basm0.
a
Similarly for the beam A G,
= W. I sin 0-B'b cos 0.
To obtain an equation between B and E, not involving the
forces at A, let us suppose the rods to become rigidly joined at
A, which will not disturb the equilibrium, nor afiect B and B' ;
BAG being now a rigid body kept at rest by B, B' acting
190
PROBLEMS.
horizontally, and its weight and the reaction of ED at A acting
in a vertical direction, we hare, taking the horizontal forces,
= R-E.
Hence, eliminating R and R between the three equations now
found, we obtain
and .•. ED = a cos ^ + J sin ^
_ a + J tan 6
~VT+t^?
_ a'JW+l4W
^/W'+W
19. Two given rods connected hy a hinge are laid across a
smooth horizontal cylinder of given radius. Determine the posi-
tion of equilibrium and the strain ujpon the hinge.
Let AC, BO (fig. 96) be the rods, resting upon the circle
whose centre is at the points P, Q. Let G, G' be their
centres of graviiy. Join OG, draw OS vertical, and upon it
drop the perpendiculars GM, G'M'. Let
OOH=e, AOO = BOO = tt>, GO=a, G'0=b,
the radius of the cylinder = r, W, W the weights of the rods.
Then if i be the altitude of the common centre of gravity of the
rods above a horizontal plane passing through the point 0,
{W+W')s=W. 0M+ W. OM'.
Now if O^cut AO'va. H, and BO pf&duced in H',
0M= 00 cose- OG . cos OHH'
rcos.d , >,,
= — r — ; a cos (<p + a),
sin^) ^^
PROBLEMS. 191
and OM'=OCcoae-CG'cQsH'
= —. — 3 — COS (A — 6);
sin <p \T / '
.: {W+W')z = (W+W')r.^?^
^ ' sm 9
- Wa cos {<j) + 0)- W'h cos (^ - 0).
Now in a position of equilibrium z, which is a function of
the two independent variables 6 and <^, must be a maximum or
a minimum ;
.-. 0={W+W')dez = -{W+W')r.^^
+ Wa sin (0 + ^) - TF'5 sin {^-6),
and = (Tr+TFV*» = -(T^+T^V--^^^^^
^ / 9 \ / sm 9
+ TFa sin {<f>+0) +W'h sin (0- ^) ;
.-. {W+W) rcoa&<f4>={Wa-W'h) coi + {Wa +W'h) cot0,
and ( W+ W')r cosec> ={Wa- W'h) tan^ + ( TFa + PF'S) tan 0.
From which we find, by subtraction,
tm_20 _ Wbj-Wa ,.
tan20~ W'b + Wa ^ ''
and by eliminating 0,
4 TFTr'aJ sin*0 - ( W+ W) ( Wa+ W'h) r tan^ + ( Pr+ Tr')V= ;
firom which ^ being found by approximation, will be known
from (1).
To find the strain upon the hinge.
Let T be the strain, exerted upon AO in the direction OT,
and upon OB in the opposite direction. To avoid the reaction
at P, which is unknown, resolve the forces, which keep AG &t
rest, parallel to AC, and take their moments about P;
.: 0=Tco3 {it- A OT) - Wcoa AEO,
and 0=T. PO sin A OT- W. PG . sin AHO.
192 PROBLEMS.
From these equations we find
iwh
cos' ASO + ^, . Bm' AHO
PG
= cos' {<i> + e) + (^ tan ^ - l) ^in" (^ + 6)
from which jP is known; and
ta.nAGT=-^. tan AHO
= (l--.tan(^)tan(^ + ^),
gives the direction in which T acts.
20. A given weight W is sustained on a given inclined plane,
partly hy friction and partly hy a power P acting in a given di-
rection. Find the greatest and least values ofF.
Let (fig. 97) be the body placed on the inclined plane AB;
let S be the reaction of the plane in a normal direction, and /jl
the coefficient of friction between the bodj and the plane : then
if the tendency of is to slide down the plane, P having its
least value, the friction /iB will act in the direction CB to pre-
vent the motion ; and therefore resolving the forces parallel and
perpendicular to AB,
= fiR + Pcos6- Wsin i,
and = ^ + Psin^-}Fcosz,
i representing the inclination of the plane to the horizon, and
the iPOB. Hence eliminating B,
-. TT7- sin i — iM cos *
P=W. ^ — . a -
cos —fismff
This is the least value of P; i.e. if Pbe less than this, (7 will
begin to slide down the plane.
If P have its greatest value, G will be on the point of moving
up the plane, and therefore the friction fiB will act down the
PEOBLEMS. 193
plane ; this will be taken account of by writing — /i for /* in the
preceding result ; consequently the greatest limit of P
_ ™ sin « + /* cos i
' cos d + fi sind'
Any value of P between these two limits will maintain equi-
librium.
CoE. The limiting values of P found above may be put
under the forms
■ffr sin(t-tan-V) ^^^ ^ sin (t + tan"^!^) _
' cos {9 + tan"' fj.) ' ' cos (0— tan"'/*) '
and, from comparing which with Art. 212, we perceive that the
least and greatest values of Pare such as would balance TFif the
inclined plane were smooth and its inclination diminished or in-
creased respectively by the angle tan~'/n.
21. To fi/nd the limiting values of "2 in the common screw
when friction acts.
Let TF be the weight sustained, « = the inclination of the
thread of the screw to the horizon ; R = the reaction perpen-
dicular to the thread, fiB = the friction along the thread : and
suppose that P has its least value. Let r = the radius of the
screw, and p the length of the arm by which P acts ; then
resolving all the forces vertically, and taking their moments
about the axis of the screw {fiB acts up the thread), since the
axis of the screw is only moveable lengthwise, by Art. 94 we
have for equilibrium
= Iico3i+ fiR sin i — W,
0= {Rsini — fiB cost) r — Pp.
By eliminating B between these equations, we find
p 1 + /x. tan I
r
E. S.
W. - . tan (i — tan"».
p
25
194 PEOBLEMS.
Hence the least value of P is the same as in a screw without
friction, the thread of which is inclined to the horizon at the
angle
^— tan"*/*.
By writing -ytt for /i, we find that the greatest value of F is the
same as in a screw without friction, the thread of which is in-
clined to the horizon at the angle
i + tan-*/A,
22. Let AC he a curve line m a vertical plane; P, Q given
weights attacked to the extremities of a string which passes over
a smooth pin at B/ to shew how to find the position of equili-
hrium. (Fig. 98.)
Take the vertical line Bx for the axis of x ; and any fixed
point ^ in it for the origin of co-ordinates : draw QM perpen-
dicular to Bx; and put AM=x, QM=y, Bp = x'; then if P
descend through a small space dx', the corresponding space
descended by Q will be dx ; and as P and Q are acted on by no
other forces than gravity, except the tension of the string and re-
action of the curve line, the virtual velocities of P and Q are dx'
and dx; and consequently, by Art. 113,
PdK^ + Qdae = ;
this is the only mechanical condition of equilibrium. The geo-
metrical nature of the machine is expressed by the equation
of the curve
y = F{x),
and the equation, (6 being the length of the string, and a de-
noting AB)
h = x+ '^{a + xf + f.
Ex. Let AC he a parabola, and B the point where the axis
intersects the directrix.
In this case y' = iax ;
.'. J = a;' + V (a + a;)' + iax ;
.'. = dx'+r-i — :; ^a-dx
(o' + 6ax + a?)*
p-'^+{a' + Gax + x'f^-
PEOBLEMS. • 195
Hence dividing by dx, and reducing the result, we find
2a Va
x = — Sa +
p2\i"
(-g
23. The weight P in Prob. 21 instead of hanging perpen-
dicularly, rests upon a given curve line AD/ to find the position
of equilibrium. (Fig. 99.)
If x, y' be the co-ordinates of P, and x, y those of Q, both
measured from B as origin, the virtual velocities of P and Q will
be respectively dx and dx ; consequently
Pdx + Qdx = 0.
To this we must join the equations of the two curves
y' = F{x'), &ndiy = F{x),
and the equation
b^'Jx'^+y'^ + V^f+f.
Ex. Let AD be a circle whose centre is in BA produced;
and AC a parabola, the directrix of which passes through B.
Then the equation of AC is
y = 4a (a; — a),
and that of AD is
y==2c(a;'-a)-(a;'-a)',
c being the radius of the circle ;
.-. b = '^x" + y" + '/^^Tf
= V2£b' {c + a)- 2ac -d'+ 'J a^ + iax - ia' ;
(c + a) dx' {x + 2a) dx
'J2x' {c + a)-'2ac — d? slx^ + ^ax—^a^
_ {c + a) dx (a;+2a) dx
~ BP ^ BQ '
Q _ doc _ x + 2a BP
_x + 2a /_5_ _ \
~ c + a '\BQ J'
196
PROBLEMS.
from which equation, BQ being known in terms of x, x may be
found.
24. Two given weights P, Q are connected by a string PAQ
which is laid across a horizontal cylinder / to find the position and
nature of the equilibrium. (Fig. 100.)
It is evident the string will lie in a vertical plane perpen-
dicular to the axis of the cylinder. Let G be the centre of the
circular section of the cylinder by this plane. Draw CA vertical,
and BGD, PM, QN horizontal : join CP, OQ. Then since the
length of the string and the radius of the cylinder are given, the
angle PCQ is known ; let it be denoted by 2a : and let (x + 0,
a — 6 represent the angles QCA, PC A; and a= CA. Then
if i be the altitude of the common centre of gravity of P and Q
above BD, we have
{P+Q)z = P.CM+Q.CN
= Pa cos {a— 6)+ Qacos (a + 0) ;
.-. (P+ Q)des = Pa sin {a.-ff)- Qa sin (a + 6),
and (P+ Q) diz = — Pa cos (a - 5) - Qa cos (a + 6)
= -{P+Q)z (1).
Now in the position of equilibrium dgs = 0, and therefore
Pa sin (a — 5) = Qa sin {ix + 6),
from which we find
/, P- Q
tan = p „ . tan a,
which gives the position of equilibrium, which is unstable, be-
cause equation (1) shews that s is then a maximum.
25. A hollow paraboloid is placed with its vertex downwards
and axis vertical; a given rod rests within it, leaning against a pin
at the focus, and having its lower end upon the parabolic surface.
Find the position of equilibriv/m. (Fig. 101.)
Let PQ be the rod, G its centre of gravity, 8 the focus of the
paraboloid, AS its axis, BA G a section of it by a vertical plane
PROBLEMS. 197
passing through the rod; a = AS, b = PG, r = P8, 6 = ASF;
through 8 draw L8 horizontal, and draw MQ vertical; let
i = MG. Then by the nature of the parabola
whence cos = 1 .
1 + cos r
Also l=GM=80cose
= {r — b) cos 6
=('-»)C^->)
2ab ,
= 2a-r h b ;
r
and d;s= — ^ (1).
In the position of equilibrium d^ = 0, and therefore
r = V2aS ;
from which the position of the rod is known. Equation (1) shews
that the altitude of G is then a minimum; and therefore the
position is one of stable equilibrium.
26. A paraboloid, formed by the, revolution of a given para-
bolic area about its axis, is placed with its convex surface upon
a horizontal plane; to find the position in which it will rest.
(Fig. 102.)
Let .4C be the axis of the parabola, inclined at an ^ ^ to
the horizontal plane : P the point on which it rests ; draw
PN vertical: then since there is equilibrium the centre of
gravity must be in the line PN (Art. 132), but it is also
in AG, the axis of the parabola, consequently it is at N.
Draw PM perpendicular to AG; let a = ^ C, 6 = BO ; then the
latus rectum = — ,
a
and.-.iW=g;
198 PEOBLEMS.
2a '
AM= — T^r- = — . cor :
ftr\ 4a
.-. AN=AM+MN
= £(cof^ + 2).
2
But because N is tlie centre of gravity, AN= - a. (Ex. 5,
Art. 177) ;
cof^ = g-2,
from which the position is known.
CoE. The least value of cot 6 is when ^ = ^ : hence when
8aV
1 = or <
3^is=or<2,
or, when a is = or < ,
the solid can only rest in equilibrium with its axis vertical.
27. Two heavy rods AC, CB connected hy a hinge at C rest
on two smooth points D, E, situated in a horizontal line : find the
position of equilibrium. (Fig. 103.)
Let G, g be the centres of gravity; and W, W the weights
of the rods AC, BC; B, B' the reactions of the points B, E
which will be at right angles to the rods, because the points on
which they rest are smooth. Join DE, and let 6, ^ denote the
angles CDE, OED; and put Oa = a, Cg = a', DE=l. The
rod -40 is kept in equilibrium by three forces, viz. its own
PROBLEMS. 199
weight at G, the reaction R at D, and the tension of the hinge C;
to avoid the last, (the magnitude and direction of which are
unknown, and are not required,) let us take the moment of these
forces about G;
.: R.DO- W.aQ,os6 = (1).
Proceeding in a similar manner with the beam OB, we find
E.EG- TF'.a'cos<^ = (2),
Again, when the equilibrium is once established, we may
suppose the hinge G to become rigid; under this hypothesis-
the rigid body AGB is kept in equilibrium by four forces,
viz. R, R', W and W. Hence resolving them vertically and
horizontally, we find
Rcose + R'cos^- W-W' = (3),
andi?sin0-^'sin^ = O (4).
These four are all the independent equations which can be
derived from the mechanical properties of the machine; there
are however two others, which express its geometrical properties,
derived from the triangle J) GE, viz.
(5) DG= . ,„ ^,. , and^C=-r— T3 — ^. (6).
From (1) and (5), we find
„_ Wa cos sin {0+<l>)
~ 6 sin ^ '
and from (2) and (6)
,_ W'a cos ^ sin (g + <^)
^~ bsind
which being substituted in (3) and (4) give
^ ^' \ bsno.(ji sm p
and = Wa sixi' cos0- W'a sin" ^ cos <^,
which two equations are sufiicient for the determination of
and <^.
200
PROBLEMS.
CoE. If the rods are equal in all respects, these two
equations become
2^ • la , j.\ /cos'^ cos''(f>^ ,„,
and = sin'' ^ cos 5 - sin" ^ cos ^ ;
the last of which gives
^ = <^ (A
or 1 = cos" + cos 5 cos ^ + cos^ {B).
Let us consider these two results separately, and
(1) When 6 = ^ equation (7) gives
COS0
~ (2 J '
whence the position of the rods is known. This position is
symmetrical with regard to a vertical line through C.
(2) The equations [B) and (7) shew that 6 and ^ are in-
terchangeable, and consequently there are, besides the symme-
trical position just found, two unsymmetrical positions of equili-
brium, similarly situated on each side of the first found position.
They may be found by means of (7) and {B).
28. A solid of any form is placed with its convex surface upon
a horizontal plane ; to find the position of equilibriu/m.
Let z =/(«, y) be the equation of the surface, referred
to three rectangular axes in the body: and let xyz be the
co-ordinates of the point in contact with the horizontal plane,
and xyz those of the centre of gravity referred to the same
axes. Then the plane on which the body stands being a
tangent plane, if a/Sy be the inclinations of the co-ordinate
axes to the horizon, -5 — a, -5 — /8, — — 7 will be the inclina-
Ji Z A
tions of the vertical through the point of contact to the co-ordi-
nate axes ; this vertical line is a normal, and therefore
PEOBLEMS.
201
sma =
sin/S =
-dji
sm7 =
1
•(!)•
But since the solid rests upon a point, the vertical through
that point must pass through the centre of gravity of the solid,
i. e. the normal at the point of contact passes through the centre
of gravity of the solid ; hence the equations of the normal give
0=^x-x + d^.{z-~s)\
^=y-y+d^- (s-i)3
These two equations, together with the equation
2 =fip^ y)
will enable us to find x, y, z ; and thence a, /3, 7 from (1).
Ex. Sw^ose the solid to he the eighth part of a sphere com--
pnihended hetween three rectangular planes: to find the position in
which it will rest with its convex surface on a horizontal plane.
Let its equation be
••• 3,^=--^, and J^ = -|.
Also from Ex. 1, Art. 177, we have
O Q Q
hence making substitution in equations (2) we obtain
x = y = z;
these two equations joined with.the equation of the surface of the
sphere give
a a a
<?^ = — 1, andc?^ = -l.
E. S.
26
202 PROBLEMS.
Consequently
sin a = sin/8 = sin7 = -t= .
29. To determine the nature of the equilibrium when a hody
of given form rests upon a given curve surface.
At the point of contact of the given body with the surface
on which it rests in equilibrium, the two surfaces will have a
common normal, which wiU. be vertical and pass through the
centre of gravity of the body. Let DAd (fig. 104) be this nor-
mal, A being the point of contact of the two surfaces BC, ic ;
and B, d being the centres of curvature of the arcs BC, he corre-
sponding to the point A ; and let (? be the centre of gravity of
the body. Let now the body be made to roll over a very small
arc AP, and thereby to come into the position b'c' ; A', G', d'
being the new positions of A, G, d; and P being the new point
of contact. By this movement the point A' will trace out a
small portion of an epicycloid, which at the very beginning of
the motion is perpendicular to the surface at A ; hence A' begins
to move along the line Ad. We suppose the displacement of the
body so small that ^' is in Ad. Draw Pj) vertical. If Pj> pass
through G' the body is still in equilibrium ; but if G' lie to the
right of i^ (as in the figure), the body when left to itself will
roll back into its original position ; and lastly, if G' lie to the
left of i^, the body will roU farther from its first position. Hence
the first position is one of stable, unstable, or neuter equilibrium
according as
A'j>is><oi = A'G'.
To express this result analytically, let p, p be the radii of
curvature DA, dA.
Then because the lines i^, DA' (for A is in the line Dd) are
parallel ;
M_A^ pjV P_
•'' DP~ A'p""^ p ~Ay
.:A'p = -Bf..
P + P
PROBLEMS. S03
Hence the equilibrium is stable, unstable, or neuter, accord-
ing as
PP
P + P
T is >< ox = AG.
COK. 1. If the surface on which the body rests be concave,
■we must account p negative in the above result.
COE. 2. If the surface be a plane, we must make p infinite,
and then since
-^--^ = P,
P+P i+fi.
P
in that case, the equilibrium will be stable, unstable, or neuter ac-
cording as
p' is >< or = ^ G'.
CoE. 3. If the lower surface of the body be a plane, we must
make p infinite, and then the result is
p is >< or = AG.
Ex. Find what segment of a paraboloid will rest in a posi-
tion of neuter equilibrium upon a spherical surface of given
radius.
Let X be the length of the axis of the paraboloid, im its latus
rectum ; and a the radius of the spherical surface. Then from
Ex. 5, p. 101, we have
and by the Differential Calculus
p = 2m,
T 2« 2ma
and .'. -7r = :
3 2«i + a
5ma
2m + a'
204, PEOBLEMS.
30. A string is stretched along a smooth curve line of any
form hy two equal forces, required the unit of pressure exerted
hy it upon the cylinder at any point. (Fig. 105.)
Let AHB be the curve line along which the string is
stretched by the two equal forces P, Q. Let EH' be a very
small arc, and at H, H' draw tangents meeting in K, and nor-
mals HO, H' 0. Join KO, and put ^ HOH' ^ Sd. The portion
of string HH' is kept in equilibrium by the tensions at H, H,
each of which is equal to P or Q; and by the reactions of
the curve line HH', which being smooth, the reaction at every
point will be in the direction of a normal. Hence the re-
sultant of all the reactions on HH' will pass through 0, and
as it must also pass through K, it acts in the line OK. Hence
by Art. 28,
resultant reaction on HH : P :: sin HKH' : sin OKH
:: sin HOH : cos KOH
:: Z9 : 1 ultimately ;
.'. resultant reaction on HH' = P. W.
Now when the arc HH' is diminished indefinitely, the pres-
sures upon it may be all considered parallel, and therefore their
resultant is equal to their sum.
Consequently the pressure upon the indefinitely small arc
HH is equal to PS^ or to
p a-rc
' rad. curv. '
and the unit of pressure (or pressure on an arc of the length
unity)
_ tension of string
~ rad. curvature
Cor. Let C, D be the points where the string leaves the
arc ; and let p be the whole pressure upon CH; and let 6 = the
angle between the normal at G and that at H; then Sp = the
pressure upon HH, and by what has just been proved
hp^ne-,
.-. _p + 0= pe.
PROBLEMS. 205
But when ^ = 0, j) = 0, and therefore C-0;
.•.p=pe.
If a be the angle included between the normals at C and D,
and p the pressure upon the whole arc OHD,
It is remarkable that this result is independent of the form of
the curve, provided it be in every part convex towards the string
in contact with it.
31. A string is stretched along a rough curve line of any
form iy two such forces that the string is on the point of moving.
Having given the coefficient of friction, find the proportion of the
forces. (Fig. 105.)
Let Q be the larger force ; <, J + S< the tensions of the string
at H, H' ; /jl the coefficient of friction ; 6, SO, a as before. Then
the pressure on MS' = tSd, and therefore the friction on SH' =
fjLtBd; but the arc MH' being pulled in opposite directions by
the forces t, t + St, the latter is prevented from producing motion
only by the friction on HH' ;
.-. t + St - t = /jLtSe ;
det
■■-1=1''
.'. log,* =^fi,e+o,
when 6 = 0, t = P, and when 6 = a, t= Q; wherefore
loge<9-log,P=/ia;
.-. Q = P^.
This result is independent of the form of the curve.
32. A unifcyrm heavy chain is laid upon a smooth arc of a
quadrant of a circle, and coincides with it; one of the hounding
radii of the quadrant heing horizontal, and the other vertical.
Find the force necessary to prevent the chain from sliding down
the arc : and compare the pressure upon the curve with the weight
of the chain. (Fig. 106.)
206
PROBLEMS.
Let F te the force whicli, acting horizontally at B, juat
prevents the chain from sliding down the quadrant. Let F, Q
be two points very near to each other ; a = AO, —A OP,
Sd = POQ, t and t+ St the tensions of the chain at P and Q;
J) = the pressure upon the arc AP, and Sp = that upon PQ ;
p = the weight of a piece of the chain of the length 1. Then
the elementary portion of -chain PQ is kept in equilibrium by
the tensions t, t + St, its own weight paSd, and the reactions of
the curve PQ ; to avoid the latter, take the moments of these
forces about the point ;
.•. = {t + St)a — ta — paS6 .acosd;
.'. dgt = pa cos ;
.\ t=pasia6+ C^
But at ^, t = 0, ^ = 0, and.-. (7 = 0;
.'. t = pa sin 0.
And at 5, < = ^ and 0=^;
.-. F=pa
= the weight of a piece of the chain, the length of
which is equal to the radius.
Again, to find the pressure upon the quadrant.
The pressure of the elementary portion PQ is due to two
causes, viz.- its own weight, and the tensions t and t + St. The
former part = paS0 sin 0, and the latter part = tS0, by Prob. 29 j
.-. Sp =paSd sin d+tS0;
.-. dep = pa sin + pa sin 0',
.-. p = — 2pa cos 0+0.
At 2,'0 = O and ^ = 0; .•.C=2pa;
.'. ^s=2joa(l — cos^),
PEOBLEMS. 207
and at J5, ^ = x and p = the whole pressure on the quadrant,
= 2pa;
press, on quad. _ 2pa _ 4
weight of chain tt tt "
33. Supposing the quadrant to be rough, to find the least
value of F which can prevent the chain from sliding off; having
given the coefficient of friction (=/*).
In this case the chain is on the point of moving towards the
point A, consequently friction acts up the quadrant.
The forces which keep PQ in equilibrium are t, t-\- Si, /iSp,
paW, and the normal reactions; to avoid the last, take the
moments of the forces about the point ;
.'. Q = {t + Zi)a — ta + fJbZpa — paZd . a cos d ;
.". det+ fidgp=pacos6 (1).
■Also as before
Sp = paZO . sin 6 + tW ;
.'. dgp = pa sin 6-\-t.
Hence substituting this value of d^p in (1), we obtain
del + /i {pa sin 6 + t) = pa cos ;
.'. det + fit = pa (cos — fj, soi. 9) .
Multiplying this equation by ^ and integrating, we find
. 2w cos 0+(l — u^ sin ..a , n
1 + /A
Now when 0=0, t = 0;
And when = J, t = F;
20S PROBLEMS.
l + jjr '^ l + fir
Cor. If the pressure be required, it may be found by inte-
grating equation (1) ;
.". t + /i^ = pa sin 6,
no constant being added, because t, p, 6 vanisli together ;
pa . „ t
.'. p='— . smo .
/* ^
Hence when = — , we have the pressure on the quadrant
^pa_F
" /* /*
_pa l—fj? pa 2pa -^
2pa , , -^
l + /i'
APPENDIX.
ON THE COMPOSITION OF TWO FORCES ACTINO ON A
POINT.
1. Since two forces which are in equilihrium must neces-
sarily be equal and opposite, two forces F^ and F.^ which do not
act in opposite directions, .must necessarily have a resultant, the
position of which we shall proceed to determine.
(1) The resultant of two forces Fj Mnd F^ acting on a point
P, is situated in the plane F^PF^.
For if it be not in that plane, it must be either above or
below. But it cannot be above ; for, any reason which would
assign it such a position might be used to assign it a similar
position below ; for these two positions are similarly situated
with regard, to the forces F^ and F^ ; there would consequently
be two resultants, which is impossible. The resultant then
cannot be situated above the plane of the forces ; and in a
similar way we may shew that it cannot be situated below, and
therefore it must be in the plane.
(2) It lies within the angle FjPF^.
For the tendency of F^ is to draw the particle P in the direc-
tion Pjfj, while that of .F^ is to draw it in the direction PF^, and
hence the real motion, which is the result of these united ten-
dencies, will not be in the direction of either, but intermediate to
both ; and therefore within the angle F^PF^ : consequently the
resultant, which. is a single force that would produce the sanae
motion, must be situated within the angle FJPF^.
2. Since F^ and F^ do in part hinder each other from pro-
ducing their whole effects, it appears that their resultant must be
less than their sum ; for their resultant Can only be equal to their
E. s. 27
210 APPENDIX.
sum when neither interferes with the other, which is not the case
unless they act in the same direction; consequently
3. If the forces Fj avd F^ are egual, their resultant R will
bisect the angle FjPF,.
For if there be a reason why FB should lie nearer to FF^^
than to PFj, there must he a similar reason why it should lie
nearer to Pi^^ than to FF^, since the forces are equal; and hence
there would he two resultants, which is impossible ; consequently
PF bisects the angle F^FF^.
4, Having thus determined the direction of the Tesultant of
two equal forces, we proceed to find its magnitude.
Let F^,f (fig. 107) be two equal forces acting on the particle
P, and i? their resultant bisecting the angle F^Ff. Since R is
less than the sum of the two forces F^ and f it is clear that
R R
■ fc, J , or its equal -^^ , is always less than 1 ; and, conse-
quently, an angle 6 inay be found such that
2^=cos0,
or P=2PjCOs^.
The angle 6 is .unknown at present, but from Art. 19 we
learn that so long as the angle F^Pf^ remains the same, 6 con-
tinues unchanged ; that is, if we have two sets of forces inclined
at the same angle with each other respectively, we shall have
B = 2F^ cos 6, and R' = 2i?"j cos 9, and therefore
R: R' :: F^: F,' (A),
that is, the resultants are proportional to the components.
Let now F^, fhe two other eqttal forces acting on P whose
resultant is also equal to R, the angles F^FF^, flf^ being each
equal to RFF^ or RPf^. Now at P apply four forces, each equal
to X, two of them respectively in the directions FF,, Pf^, and the
•other two in the direction PR ; and let them be of such magni-
APPENDli. 211
tude, that F^ may be the resultant of the one in the direction
PF^ and one in the direction PB. Then, since these two contain
the same angle as F^ and f^, and F^ is their resultant,
F^ = Ix cos Q.
Also, if we substitute instead of F^ and f^, their components,
we may consider R as the resultant of the forces x, x, x, and x ;
of which two act in the same direction as M ; and, consequently,
P—2x is the resultant of the two x, x, which act in the direc-
tions PF^, Pf^ ; and since, by hypothesis, B is the resultant of
F^ and f^, which act in the same directions as x, x,
.: B : B-^x :'. F^: x, from (A) ;
• ^-^ 2
But i2= 2i?; cos 5= 2 . 2a;cos^ . cos 6 = lx cos'^;
.-. 4=4cos^^-2
= 2(2cos=e-l)
= 2 cos 2^;
.-. jB=2i^jCos2^.
It appears then, that in the formula
B = 2^1 cos e,
if we double the angle at which the forces are inclined, we must
also double 0.
We will now suppose, that when the angle at which the
forces act is a multiple n, or any inferior multiple, of FJPf^, it is
true that in the same formula the corresponding equimultiple
of is to be taken ; so that
B = 2i^„ cos n0 = 2F^^ cos (w - 1) ^ = . . . = 2F^ cos 6.
. Apply (fig. 108) at P, as before, four forces in the direc-
tions PF^, PF^_^, Pf^^, and Pf„_^ respectively, each of such a
212 APPENDIX.
magnitude x that F may Ije the resultant ,of the two in the direc-
tions -Pi^„+,, Pi^„_i, and/„ of the other two ;
.\ F„ = 2a; cos d*
But if, instead of the forces F^, /„, we substitute their four
components, we may consider B as the resultant of the forces
03, X, X, and x, of which two acting in the directions PF^^ , Pf„_i
will have 2a; cos (w — 1) for their resultant in the direction FB,
and consequently B — 2x cos {n — l)d is the resultant of the other
two which act in the same directions as F^^ and_^j; consequently,
from (A),
B : ^-2a;cos(M-l)^ :: F^i : x;
B -H „ / ,x /I
••• ^^ = --2cos(«-l)5
= 2 COS cos «0 — 2 sin sin w0.
a;
But B = 2i^„ cos n0 = 4cc cos cos n0 ;
.". -p— = 4 cos 5 cos w^ — 2 cos cos «^ — 2 sin sin n0
= 2 (cos cos «0 — sin sin w^)
= 2 cos (w + 1) ;
.•.'5 = 2i?'^,cps(»i+l)5.
Hence the formula is true for a multiple {n + 1) if it he
true for n and all inferior multiples: but it has been shewn
to be true for 2 and 1, and consequently it is true for multiples
3, 4, 5, 6, ... and generally, by induction, for any multiple
whatever.
It appears then, that as we mcrease the angle at which two
equal forces {F, f) act, we must increase the angle in the same
proportion, and then, that the formula
B = 2i^cos
For the lF.^.t PFn-i = /Fj Pf„ (fig. 107).
APPENDIX. 213
still holds good. This, however, supposes the angle between
the forces to be a multiple oi F^Pf^ (fig. 107), which may not
happen to be the case ; but by taking the original angle Fj^Pf^
exceedingly small, we may find a multiple of it which shall differ
from FPfsi proposed angle by less than any assignable quantity.
It is evident then, that FFf and have an invariable ratio to
each other, so that if FPf— 2^, then
6
— = constant quantity = o suppose ;
.-. i2 = 2^cosc^.
To determine the value of c, we observe that if i^ and /act at
an angle tt, or are opposite to each other, (in which case ^ = -k")
they have no resultant ;
.-. = 2Fcos^,
.'. cos — = 0.
Now none but angles which are odd multiples of — have
their cosines = ;
.'. c = an odd integer, = 1 as we shall shew.
TT
For if c is not = 1, let the angle FPf be such that ^ = ^ >
which is therefore less than a right angle, and then
R = 2ii'cos cj> = 2i^cos ^ = 0.
But since the angle FPf is, in this case, = - , and therefore
less than tt, the resultant cannot be == 0, which is absurd, and
consequently c = I. We arrive therefore at the general result,
that if ii^,/be two equal forces acting on a particle, and inclined
to each other at the angle 2<^, their resultant R is inclined to
^14 APPENDIX.
each of them at the angle ^, and its magnitude is determined by
the equation
jB = 2^cos^.
5. It will be immediately obvious that, since the forces
F and / are perfectly equal and similarly situated with respect
to PR, they contribute equally to the resultant R ; and, conse-
quently the efficiency of each in the direction PR is equal to ^R,
or Fcos<f).
6. To determine the magnitude and direction of the resultant
of any two forces acting on a particle.
Let F, f (fig.*109) be the two forces acting on the particle
P; R their resultant, perpendicular to which draw iPJf ; let
a, /3 denote the angles FPR, fPR respectively, and ^ the angle
jFPf between the forces. Then the efficiencies of F and/, in the
direction PR, are respectively ii^cos a, /cos/S, the sum of which
must be equal to R, since the efficiency of R is equivalent to the
united efficiencies of F and f in any proposed direction, because
R is their resultant;
.-, 5=i?'cosa+/cos/3 (1).
Now the efficiency of R in the direction PL perpendicular
to itself = R cos 90° = ; and the efficiency of F in the direction
PL = J' cos FPL, and that of /in the same direction = f cos f PL ;
.: = Fcos FPL +f cos f PL,
or = Fcosf^ - a) +/C0S ^l + ^)' ,
or = l'sina-/sinj8 (2),
and by squaring equations (1) and (2), we have
R' = F' cos" a + 2Ff cos a co^)8 +f cos" /S,
= i^^'sin'' a-2Ffsin a sinj8+/'sin'iS;
and adding these together,
R^^F'' + 2 Ff (cos a cos jS -'sin a sin /8) +f\
APPENDIX. 215
But tecause </> = a + j8 ;
/. cos (^ = cos a cos |8 — sin a sip /&; and, consequently,
E' = F' + 2Ffcos<f)+f.
This equation shews that the diagonal of a par3,llelogram
represents the magnitude of the resultant of two forces, which
are themselyes represented in magnitude and direction hy the
sides : and equation (2) shews that the same diagonal also repre-
sents the direction of the resultant.
MISCELLANEOUS PROBLEMS.
1. Two given weights are suspended from the ends of a
bent lever, the anns -of which are given, and include a given
angle; find the position of equilibrium.
2. A bent lever of uniform thickness rests with its shorter
arm horizontal. But if the length of this arm were doubled lie
lever would rest with the other arm horizontal. Compare the
. lengths of the arms, and find their inclination.
3. Two forces act at angles oyS upon the arms a, & of a
straight lever which Is not attached to its fulcrum. Shew that
if there be equilibrium a : b :: tan yS : tan a.
4. The beam of a false balance being uniform, shew that
the lengths of the arms are respectively proportional to the
differences between the true and apparent weights of a given
substance.
5. A beam of oak 30 feet long balances upon a point 10 feet
from one end : but when a weight of 10 lbs. is -suspended at the
thin end, the prop must be moved 2 feet to preserve equilibrium.
Find the weight of the oak.
6. Two equal forces act in opposite directions along two
opposite sides of a parallelogram, and a third force along the
diagonal. Find the force which will keep them in equilibrium.
7. If forces proportional to the sides of a polygon be applied
in the plane of the figure at the middle points of the sides and
perpendicular thereto, they will balance.
8. A given body Is supported on an inclined plane, first by
a power parallel to the base, and then by a power parallel to the
plane. Compare the pressures on the plane in the two cases.
MISCELLANEOUS PROBLEMS. 217
9. A rope of given length is used to pull down a vertical
pillar ; at what height from the base of the pillar must it be fast-
ened that a given force pulling it may be most efficacious ?
10. A weight P hangs vertically by a string from a fixed
pomt A ; a string P£W being now fastened to Pis passed over
a fixed pulley £ (so that BP is horizontal) and supports a weight
W. Find how much this will draw AP from the vertical.
11. C, Z) are two smooth pegs, and ACDB is a heavy cir-
cular arc, which passes over one peg and under the other ; find
the position of equilibrium.
12. A given sphere rests between two given inclined planes ;
find the pressure upon each.
13. Two weights support each other on two given inclined
planes which have a common verteS, by means of a string pass-
ing over the vertex ; find the proportion of the weights.
14. A given cone is placed with its base on an inclined
plane, the coefficient of friction for which is known : determine
whether, upon increasing the inclination of the plane, the cone
will tumble or slide.
15. A weight is suspended from one extremity of a string
which passes over two fixed pulleys and through a ring at its
other extremity ; find the position of equilibrium.
16. A given beam rests vdth its lower end on a smooth
horizontal plane, and its upper end on a given inclined plane ;
find the force which must act at the foot of the beam to prevent
sliding.
17. Two given heavy particles being connected by an in-
flexible rod of given length are placed within a hemispherical
bowl; find the position of equilibrium, and the compressing
force upon the rod.
18. A rigid rod AB is moveable in a vertical plane about a
fixed hinge A, the end B leans against a smooth vertical wall.
Find the pressures on the wall and hinge.
E. s. 28
218 MISCELLANEOUS PROBLEMS.
19. A beam of given length and weight is placed with one
end on a Tertical, and the other on a horizontal plane ; find the
force necessary to keep it at rest, and the pressures on the two
planes.
20. A and B are two given points in a horizontal line, to
•which are fastened two strings AG, BCWoi given lengths ; the
string BOW passes through a ring attached to the string AG,
and to it is fastened a given weight W; find the position in
which the ring will rest.
21. AG, BGD are two given beams moveable in a vertical
plane about hinges A, B'vo.& horizontal plane. BD the longer
leans upon the end G oi AG the shorter. Find the position of
equilibrium.
22. If a rod rest in equilibrium with its ends on two smooth
inclined planes, the intersection of the planes must be a horizon-
tal line.
23. A beam has a ring at one extremity which moves up
and down a vertical rod. Find the position of the beam when it
rests upon the arc of a circle a diameter of which coincides with
the rod;
24. The upper end of a given rod rests against a smooth
vertical plane, and the lower end is suspended by a given string
fastened to a point in the plane ; find the position of equilibrium.
25. A given uniform rod passing freely through an orifice
in a vertical plane rests in equilibrium with one end upon a
given inclined plane ; find its position.
26. A heavy beam leans against an upright prop ; the lower
end of the beam rests upon the horizontal plane and is prevented
from sliding by a string tied to the bottom of the prop ; required
the tension of this string.
27. Out of a square it is required to cut a triangle such that
the remaining figure may have its centre of gravity where the
vertex of the triangle was.
MISCELLANEOUS PROBLEMS. 219
28. If the sides of a triangle taken in order be cut propor-
tionally, the triangle formed by joining the points of division
will have the same centre of gravity as the original triangle.
29. Find the distance of the centre of gravity from the
angular point of a uniform bent lever whose arms and the angle
which they include ai-e given.
30. A solid composed of a cone and a hemisphere of equal
bases, placed base to base, rests with the convex surface of the
hemisphere upon a horizontal plane, and the axis of the cone in
an inclined position; compare the dimensions of the cone and
hemisphere.
31. Determine the point in its curve surface on which a
semi-paraboloid will rest on a horizontal plane.
32. A solid generated by the revolution of a quadrant of an
ellipse about its major axis, is placed upon a horizontal plane,
with its axis in an oblique position ; determine the position of
equilibrium.
33. An ellipsoid rests on a horizontal plane on the extremity
of its mean axis ; shew how to estimate the stability with regard
to a slight displacement in any direction. Define the direction
which distinguishes between stable and unstable equilibrium.
34. The centre of gravity of three weights a. {w — a)",
b.{w — /8j', c . {w — <yf, whatever be the value of w, will be
situated in a line of the second order to which the lines joining
the centres of gravity of the weights are tangents.
35. If a hemisphere and paraboloid of equal bases and
similar materials have their bases cemented together, the whole
solid will rest on a horizontal plane on any point of the sphe-
rical surface if the altitude of the paraboloid : the radius of the
hemisphere :: \/3 : V2.
36. Three uniform beams AB, BC, CD, of the same thick-
ness, and of lengths I, 2l, I respectively, are connected by hinges
at B and C, and rest on a perfectly smooth sphere, the radius
220 MISCELLANEOUS PROBLEMS.
of whicli = 2l, SO that the middle point of BG, and the extre-
mities A, D are in contact with the sphere; shew that the
91
pressure at the middle point of 50'=- of the weight of the
beams.
37. A sphere of given weight and radius is suspended hj
a string of given length from a fixed point, to which point also
is attached another given weight by a string so long that the
weight hangs below the sphere ; find the angle which the string,
to which the sphere is attached, makes with the vertical.
38. Two given beams AG, BDG lean against each other
in a vertical plane ; and their ends A, B resting on a smooth
horizontal plane are prevented fi-om sliding by a string AD,
which is fastened to the beam AG at A, and the beam BDG
at D. Find the tension of the string.
39. A cylinder, with its base resting against a smooth ver-
tical plane, is held up by a string fastened to it at a point of
its curved surface whose distance from the vertical plane is h.
Shew that h must be >b — 2a tan and < b, where 25 is the
altitude of the cylinder, a the radius of the base, and d the angle
which the string makes with the vertical.
■'o
40. A flat board in the form of a square is supported upon
two props with its plane vertical; determine its positions of
equilibrium, friction being neglected, and the distance between
the props being equal to half a side of the square.
41. Determine the position of equilibrium of a uniform rod,
one end of which rests against a plane "perpendicular to the
horizon, and the other on the interior surface of a given hemi-
sphere.
42. If the sides of a triangle ABG be bisected in the points
D, E, F; then the centre of the circle inscribed in the triangle
DEF is the centre of gravity of the perimeter of the triangle
ABG.
43. Three equal rods, loosely connected together by one
extremity of each, have their other extremities placed upon a
MISCELLANEOUS PEOBLEMS. 221
rough horizontal plane at the angular points of a given equi-
lateral triangle. A smooth heavy ring ia then placed on the
rods ; find tlie coeflficieut of friction between the rods and the
plane that the machine may just he on the point of falling.
44. A cone and sphere of given weights support each other
between two given inclined planes, the cone resting on its base
on one of the planes ; determine what must be the vertical angle
of the cone, that the equilibrium may subsist.
45. A given cylinder with its axis horizontal is held at
rest on a given rough inclined plane by a string coiled round
its middle and then fastened to the top of the plane ; find the
position of equilibrium.
46. A given weight is placed upon a rough horizontal
plane ; required the magnitude and direction of the least force
which will be able to move it.
47. The resultant and sum of two forces are given, and also
the angle which one of them makes with the resultant; it is
required to determine the forces and the angle at which they act.
48. A circular hoop is supported in a horizontal position,
and three weights of 4, 5, and 6 pounds respectively are sus-
pended over its circumference by three strings fastened together
in a knot. Shew that the knot must be in the centre of the
hoop, and find what must be the positions of the strings so that
they may sustain one another.
49. Four beams, AB, BG, CD, DA (fig. 27) connected by
hinge joints, have the opposite comers connected by two elastic
strings A C, BD. Shew that
AE.EG BE. ED
tension oi AG : tension of BD
AG ' BD
50. A uniform straight rod rests with its middle point upon
a rough horizontal cylinder, their directions being horizontal
and perpendicular to each other. Find the greatest weight
which may be attached to one end of the rod without causing
it to slide oif the cylinder.
222 MISCELLANEOUS PROBLEMS.
51. Two equal uniform beams connected by a binge joint
^are laid across a smooth horizontal cylinder of given radius.
Find their inclination to each other when in equilibrium,
52. A particle is placed on the surface of an ellipsoid, and
is attracted towards the principal planes by forces which are
respectively proportional to its distance from them; find the
conditions of equilibrium.
53. Prove the equality of the power and weight in Rober-
val's balance by couples ; and find the strains upon the joints
and pins.
54. A particle is placed on the arc of a given parabola, and
is acted on by gravity parallel to the axis, and a force perpen-
dicular to it which is proportional to the distance of the particle
from the axis ; find the position of equilibrium.
55. If three parallel forces acting at the angular points
A, B, C of a plane triangle are respectively proportional to the
opposite sides a, b, c ; prove that the distance of the centre of
parallel forces from A
2bc A
' = , — cos—.
a+h+c 2
56. A ladder rests with its foot on a horizontal plane, and
its upper extremity against a vertical wall; having given its
length, the place of its centre of gravity, and the ratios of the
friction to the pressure both on the plane and on the wall ; find
its position when in a state bordering upon motion.
57. If a lever, kept at rest by weights P, Q, suspended
from its arms a, b, so that they make angles a, ^,, with the
horizon, be turned about its fulcrum through an angle 20, prove
that the vertical spaces desci-ibed by P and Q, are to one an-
other as a cos (a 4- ^) : b cos {fi—d); and thence deduce the
equation of virtual velocities.
58. If /S and D represent respectively the semi-sum and
semi-difference of the greatest and least angles, which the direc-
tion of a power supporting a weight on a rough inclined plane
MISCELLANEOUS PROBLEMS. 223
may make with the plane, and ^ be the least elevation of the
plane when a body would slide down it; prove that the cosine of
the angle, at which the same power being inclined to a smooth
plane of the same elevation would support the same weight,
COS 8 ,„ ,,
cos ^ ^ ^'
59. A roof A CB consisting of beams which form an iso-
sceles triangle with its base AB horizontal, supports a given
weight at C; find the horizontal force at A. Why must a
pointed arch carry a heavy weight at its vertex ?
60. Four equal uniform beams connected by joints are sym-
metrically placed in a vertical plane, in the form of a roof;
shew that if the extremities be in a horizontal line, and 6, ^ be
the inclinations of the beams to the horizon, tan = 3 tan ^.
61. A beam ^B, capable of motion in every direction about
a fixed ball and socket joint at A, rests with its end B against
a rough vertical plane ; determine the extreme positions of equi-
librium.
62. In the last question suppose the end B rests against
a rough inclined plane; determine the extreme positions of
equilibrium.
63. Three weights are suspended from the angles of an
isosceles triangle, whose plane is vertical and which is supported
by a horizontal axis, passing through its centre of gravity, about
which it is capable of revolving. Determine its positions of
equilibrium, the two weights suspended from the extremities of
the base being equal, and each greater than the third: and
shew in each case whether the equilibrium will be stable, un-
stable, or neutral.
64. A uniform rod, whose length is a, moveable freely in
a vertical plane about a hinge at one extremity, is. attracted by
a force varying as D^, and acting from a centre at a height a
directly above the hinge ; find the position in which it will rest,
and the nature of the equilibrium, supposing that the attractive
force on the hinge is \g.
224 MISCELLANEOUS PROBLEMS.
65. A hollow cylinder stands upon a smooth horizontal
plane, and a light rod of given length, being in the same vertical
plane with the axis of the cylinder, passes over the upper edge
and rests against the interior surface. A given weight is attached
to the other extremity of the rod, and the cylinder is just on the
point of turning over. Determine its weight.
66. A cylinder is laid upon two equal cylinders all in
parallel positions, and the lower ones resting in contact with each
other upon a rough horizontal plane ; find the relation between
the coefiicients of friction between the cylinders, and the co-
efficient of friction between a cylinder and the plane, that all the
points of contact may begin to slip at the same instant.
67. Determine the conditions of equilibrium of a material
point situated in an indefinitely thin bent tube of any form
and acted upon by any number of forces.
68. A chain of uniform density is suspended at its extre-
mities by means of two tacks in the same horizontal line at a
given distance from each other ; find the length of the chain so
that the stress upon either tack may be equal to the chain's
weight.
69. A uniform chain is suspended from two tacks in the
same horizontal line at a given distance from each other. Find
the length of the chain that the stress on the tacks may be
the least possible.
70. A cylinder rests with the centre of its base in contact
with the highest point of a fixed sphere, and four times the
altitude" of the cylinder is equal to a great circle of the sphere ;
supposing the surfaces in contact to be rough enough to prevent
sliding in all cases, shew that the cylinder may be made to rock
through an angle 90", but not more, without falling.
^ 71. A man runs round in the circumference of a given
circle with a given velocity ; determine the inclination of his body
to the horizon.
MISCELLANEOUS PROBLEMS. 225
72. One end of a heavy rod can turn in every direction
about a fixed point. The other end rests on the upper surface
of a rough plane, (coefBciei^t of friction fi) which is inclined to
the horizon at an angle a. If /3 be the angle which the beam
makes with the plane, prove that the rod will not rest in every
position, unless cofa be not less than —^ + tan^/S.
73. A chain suspended at its extremities from two tacks
in the same horizontal line forms itself into a cycloid; prove
that the density at any point cc sec' {^0), and the weight of the
corresponding arc <x tan {^6), 6 being the arc of the generating
circle measured from the vertex.
74. A weight W is suspended from a point P of a uniform
catenary APA'. and 0' are the lowest points of two uniform
catenaries, of which AP and AP are parts. Shew that W is
equal to the difference or sum of the weights of the portions
OP, O'P of the catenaries, according as AP and AP are one or
hoik less than a semi-catenary.
75. If a chain acted on by gravity hang in the form of
the curve whose equation is sec - = e" , shew that at every point
the density or thickness is proportional to the tension.
76. A uniform catenary of given length is suspended from
two given points at the same height, and is nearly horizontal ;
in consequence of an expansion of its materials the vertex of the
catenary is observed to have descended through a small given
altitude ; find the increase of the length of the catenary, supposing
its expansion to have been uniform throughout.
77. A uniform elastic string being of such a length that
when it hangs vertically, if an equal quantity were appended
to the lowest point it would stretch it to twice that length ; what
weight must be appended at the middle point that the increase
of length may be three quarters of the original ?
78. A given heavy elastic ring is passed over the vertex of
a smooth vertical cone, and descends by its own weight ; required
the position of equilibrium.
E. s. 29
226 MISCELLANEOUS PROBLEMS.
79. A uniform heavy elastic string (natural length a) is
stretched by forces applied at its ends, and then, being laid
upon a rough inclined plane, is suffered slowly to contract itself.
Shew that a point of- the string, the natural distance of which
from the upper end is
a f tan a\
2 I V/
will not be affected by friction, a is the inclination of the plane.
80. Find the form of a uniform chain suspended from any
two points on the surface of an upright cone, and resting on the
curve surface. Find the tension when it becomes a horizontal
circle.
81. A given uniform rod is placed within a given rough
hemispherical bowl ; find the limiting positions of equilibrium.
82. If a right-angled triangle be supported in a horizontal
position by vertical threads fasteiied to its angular points, each
of which can just bear an additional tension of lib., determine
within what portion of the area a weight less than 3 lbs. may be
placed without destroying the equilibrium.
83. Find the magnitude of the horizontal strain which
a door exerts on its hinges ; shew that the vertical strain on
each hinge is indeterminate.
84. A beam, having one end on a vertical, and the other
on a horizontal plane, is kept at rest by a string connecting
its centre of gravity with the intersection of the planes. Find
the tension of the string ; and explain the result when the beam
is uniform.
85. A cycloid is placed with its axis vertical; a weight is
supported upon its arc by an elastic string, the natural length
of which is given, and one end of which is fastened to the top
of the cycloid ; find the position of equilibrium.
86. Three equal spheres are placed in contact upon a rough
horizontal plane. K another equal sphere, placed upon them,
just causes them to separate, what is the coeflficient of friction ?
MISCELLANEOUS PROBLEMS. 227
87. An elastic chain is laid upon a smooth inclined plane,
one end being made fast to the top of the plane. The natural
length of the chain is equal to the length of the plane ; find
how much of the chain will hang down off the plane when
there is equilibrium.
88. A string binds tightly together two smooth cylinders of
given radii. Compare the mutual pressure between the cylinders
with the whole pressure of the string upon them.
89. Three equal smooth spheres are placed in mutual contact
on a smooth horizontal plane, and are bound together by an
elastic string in a plane containing their centres, the string not
being stretched; another equal sphere is then placed upon
them, and sinks tiU its lowest point is on a level with their
centres. Find the elasticity of the string.
90. A string passing underneath a heavy pulley has its ends
fastened to two points in a horizontal plane, the distance between
the points being equal to the diameter of the pulley. Suppose
the string to become elastic, and the pulley to be rough, find
how far the pulley wiU sink below its first position.
91. When any number of forces act on a body, shew that
the plane on which the sum of the projections of the moments of
the forces about a fixed point is a maximum, is perpendicular to
the plane with respect to which this sum is 0.
92. Assuming that if S^, Sg-, hr be the virtual velocities of
three forces P, Q, R which keep a point at rest,
P^+QZq+Rhr = Q,
in whatever direction the virtual motion of the point takes place ;
prove that the forces are proportional to the sides of a triangle
drawn in their directions.
93. If A, B, O represent the moments of a force round each
of three rectangular axes which meet in a point, and a, /8, 7 be
the angles which a straight line through the point of intersec-
tion makes with each axis, the moment of the force round this
line is A cosa + 5coSj8 + Gcoay.
228 MISCELLANEOUS PROBLEMS.
94. V Three forces act on a point in directions respectively
perpendicular to three rectangular co-ordinate planes, and each
varying as the co-ordinate to which it is parallel; shew that
there are two planes, in either of which if the point be situated
the resolved part of the whole force,' which is parallel to the
plane, tends to the origin and varies as the distance of the point
from it.
CAMBBIDGE: PBINTBD by C. J. clay, M.A. at the UWIVBR3ITY PRESS.
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